 # Circular Motion

## Concept Of Physics

### H C Verma

1   11. Find the acceleration of the moon with respect to the earth from the following data : Distance between the earth and the moon = 3.85 x 10 5 km and the time taken by the moon to complete one revolution around the earth = 27.3 days.

##### Solution :

Distance between Earth and Moon $\\$ $T$ = $27.3$ $days$ =$24$x$3600$x$(27.3)sec$= $2.36$x$10^6 sec$ $\\$ v = $\frac{2\pi f}{T}$ =$\frac{ 2\times3.14 \times3.85\times10^8}{2.36 \times 10^6}$ = $1025.42 m/sec$ $\\$ $a = \frac{v^2}{r}$ =$\frac{(1025.42)^2}{64 \times 10^5}$ = $0.00273m/sec^2$=$2.73\times10^-3m/sec^2$

2   2. Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

##### Solution :

Diameter of earth = 12800km $\\$ Radius R = 6400km = 64 x 10^5 $\\$ V = $\frac{2 \pi R}{T}$ = $\frac {2\times3.14 \times64\times10^5}{24\times3600}$ m/sec=465.185 $\\$ a = $\frac{V^2}{R}$= $\frac{(46.5185)^2}{64\times10^5}$ = $0.0338m/sec^2$

3   3. A particle moves in a circle of radius 1.0 cm at a speed given by v = 2.0 t where v is in cm/s and t in seconds. (a) Find the radial acceleration of the particle at t = 1 s. (b) Find the tangential acceleration at t = 1 s. (c) Find the magnitude of the acceleration at t = 1 s.el.

##### Solution :

V = 2t, r= 1cm $\\$ a) Radial acceleration at t= 1 sec $\\$ a= $\frac{V^2}{R}$ = $\frac{2^2}{1}$ = $4cm/sec^2$ $\\$ b) Tangential acceleration at t =1 sec $\\$ a= $\frac{dv}{dt}$ = $\frac{d}{dt}$(2t) = $2cm/sec^2$ $\\$ c)Magnitude of acceleration at t=1 sec $\\$ a=$\sqrt4^2+\sqrt2^2$ = $\sqrt20 cm/sec^2$

4   4. A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible ?

##### Solution :

Given that m = 150kg. $\\$ v=36km/hr = 10 m/sec, r=30m $\\$ Horizontal force needed is $\frac{mv^2}{r}$ = $\frac{150\times(10)^2}{30}$ = $\frac{150\times100}{30}$ = 500N

5   5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking ?

##### Solution :

In the diagram $R cos\theta =mg ..(i)$ $\\$ $R sin\theta= \frac{mv^2}{r}$ $\\$ Dividing equation (i) by equation (ii) $Tan\theta = \frac{mv^2}{rmg}=\frac{v^2}{rg}$ $\\$ $v=36km/hr = 10m/sec, r=30m$ $\\$ $Tan\theta =\frac{v^2}{rg} = \frac{100}{30\times10} =(1/3)$ $\\$ $\Rightarrow \theta = tan^-1 (1/3)$

6   6. A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle of banking ?

##### Solution :

Radius of Park = r = 10m $\\$ speed of vehicle = 18km/hr =5m/sec $\\$ $Angle of banking tan \theta = \frac{v^2}{rg}$ $\\$ $\Rightarrow \theta = tan^-1 \frac {v^2}{rg} = tan^-1 \frac{25}{100} = tan^-1 (1/4)$

7   7. If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that a scooter going at 18 km/hr does not skid ?

##### Solution :

The road is horizontal (no banking) $\\$ $\frac{mv^2}{R} =\mu N$ $\\$ $\frac{mv^2}{R}=\mu mg \hspace{1cm} v=5m/sec, R =10m$ $\\$ $\Rightarrow \frac{25}{10} = \mu g \Rightarrow \mu = \frac{25}{100} = 0.25$ $\\$

8   8. A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used ?

##### Solution :

$Angle\hspace{1cm} of\hspace{0.5cm} banking = \theta = 30 ^{\circ}$ $\\$ Radius =r = 50m $\\$ $tan\theta = \frac{v^2}{rg} \Rightarrow tan 30^{\circ} = \frac{v^2}{rg}$ $\\$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{v^2}{rg} \Rightarrow = v^2 = \frac{rg}{\sqrt{3}} = \frac{50\times10}{\sqrt{3}}$ $\\$ $\Rightarrow v = \sqrt\frac{500}{\sqrt{3}} = 17m/sec.$ $\\$

9   9. In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coulomb attraction. In the ground state, the electron goes round the proton in a circle of radius 5.3 x 10 -11 m. Find the speed of the electron in the ground state. Mass of the electron = 9.1 x 10^ -3 kg and charge of the electron = 1.6 x 10 -19 C.

##### Solution :

Electron revolves around the proton in a circle having proton at the centre. $\\$ Centripetal force is provided by coulomb attraction. $\\$ $r=5.3 \rightarrow t 10^-11m \hspace{1cm} m=mass\hspace{0.5cm}of \hspace{0.5cm}electron=9.1\times10^-3kg$ $\\$ $charge\hspace{0.5cm}of\hspace{0.5cm} electron = 1.6\times10^-19c$ $\\$ $\frac{mv^2}{r} = k\frac{q^2}{r^2} \Rightarrow v^2 = \frac{kq^2}{rm} =\frac {9\times10^9\times1.6\times1.6\times10^-38}{5.3\times10^-11\times9.1\times10^-31}=\frac{23.04}{48.23} \times 10^{13}$ $\\$ $v^2= 0.477\times10^13=4.7\times10^{12}$ $\\$ $\Rightarrow v = \sqrt{4.7\times10^{12}}= 2.2\times10^6 m/sec$ .

10   10. A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the minimum speed the stone can have at the highest point of the circle.

##### Solution :

At the highest point of a vertical circle $\\$ $\frac{mv^2}{R}= mg$ $\\$ $\Rightarrow v^2 = Rg \Rightarrow v = \sqrt{Rg}$

11   11. A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1 g sticking at the outer end of a blade. How much force does it experience when the fan runs at full speed ? Who exerts this force on the particle ? How much force does the particle exert on the blade along its surface ?

##### Solution :

Diameter of the ceiling fan= 120cm $\\$ $\therefore Radius = r =60cm =0/6m$ $\\$ Mass of particle on the outer end of the blade is 10. $\\$ $n = 1500 rev/min =25 rev/sec$ $\\$ $\omega = 2\pi n = 2 \pi \times25 = 157.14$ $\\$ $Force\hspace{0.25cm}of\hspace{0.25cm}the\hspace{0.25cm}particle\hspace{0.25cm}on\hspace{0.25cm} the\hspace{0.25cm} blade = Mr\omega^2=(0.001)\times0.6\times(157.14) = 14.8N$ $\\$ The fan runs at full speed in a circular path. This exerts the force on the particle (inertia). $\\$The particle also exerts a force of 14.8N on the blade along the surface.

12   12. A mosquito is sitting on an L.P. record disc rotating on a turn table at 33$\frac{1}{3}$ per minute. The distance 3 of the mosquito from the centre of the turn table is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than It 2/81. Take g =$10 m/s^2.$

##### Solution :

A pendulum is suspended from the ceiling of a car taking a turn $\\$ $r=10m, v=36km/hr=10m/sec\hspace{0.1cm} g=10m/sec^2$ $\\$ From the figure Tsin$\theta={mv^2}{r} ..(i)$ $\\$ T$cos\theta=mg$ ..(ii) $\\$ $\Rightarrow \frac{sin\theta}{cos\theta}=\frac{mv^2}{mg}\Rightarrow tan\theta=\frac{v^2}{rg} \Rightarrow \theta=ta{n}^-1\frac{v^2}{rg}$ $\\$ $=tan^-1\frac{100}{10\times10}=tan^-1(1) \Rightarrow \theta = 45^{\circ}$

13   13. A simple pendulum is suspended from the ceiling of a car taking a turn of radius 10 m at a speed of 36 km/h. Find the angle made by the string of the pendulum with the vertical if this angle does not change during the turn. Take g= $10 m/sec^2$.

##### Solution :

A pendulum is suspended from the ceiling of a car taking a turn $\\$ $r=10m, v=36km/hr=10m/sec\hspace{0.1cm} g=10m/sec^2$ $\\$ From the figure Tsin$\theta={mv^2}{r} ..(i)$ $\\$ T$cos\theta=mg$ ..(ii) $\\$ $\Rightarrow \frac{sin\theta}{cos\theta}=\frac{mv^2}{mg}\Rightarrow tan\theta=\frac{v^2}{rg} \Rightarrow \theta=ta{n}^-1\frac{v^2}{rg}$ $\\$ $=tan^-1\frac{100}{10\times10}=tan^-1(1) \Rightarrow \theta = 45^{\circ}$

14   14. The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at this instant.

##### Solution :

Bob has a velocity of 1.4m/sec,when the string makes an angle of 0.2 radian. $\\$ m=100g=0.1kg r=1m, v=1.4m/sec $\\$ From the diagram, $\\$ T - mgcos$\theta$=$\frac{mv^2}{R}$ $\\$ $\Rightarrow T = \frac{mv^2}{R} +mgcos\theta$ $\\$ $\Rightarrow T = \frac{0.1\times{1.4}^2}{1} + 0.1\times9.8\times[1-\frac{{\theta}^2}{2}]$ $\\$ $\Rightarrow = 0.196 + 9.8 \times [1-\frac{({2})^2}{2}] (\therefore cos \theta = 1-\frac{{\theta}^2}{2} for small \theta)$ $\\$ $\Rightarrow = 0.196 + (0.98) \times (0.98)= 0.916 + 0.964= 1.156N \approx 1.16N$ $\\$

15   15. Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with the vertical. Find the tension at this instant. You can use cos$\theta$ = 1- $\theta^2 /2\hspace{0.2cm} and\hspace{0.2cm} sin\theta$ = 0 for small$\hspace{0.1cm}$$\theta$. $\\$

##### Solution :

Bob has a velocity of 1.4m/sec,when the string makes an angle of 0.2 radian. $\\$ m=100g=0.1kg r=1m, v=1.4m/sec $\\$ From the diagram, $\\$ T - mgcos$\theta$=$\frac{mv^2}{R}$ $\\$ $\Rightarrow T = \frac{mv^2}{R} +mgcos\theta$ $\\$ $\Rightarrow T = \frac{0.1\times{1.4}^2}{1} + 0.1\times9.8\times[1-\frac{{\theta}^2}{2}]$ $\\$ $\Rightarrow = 0.196 + 9.8 \times [1-\frac{({2})^2}{2}] (\therefore cos \theta = 1-\frac{{\theta}^2}{2} for small \theta)$ $\\$ $\Rightarrow = 0.196 + (0.98) \times (0.98)= 0.916 + 0.964= 1.156N \approx 1.16N$ $\\$

16   16. Suppose the amplitude of a simple pendulum having a bob of mass m is $\theta_{\circ}$ . Find the tension in the string when the bob is at its extreme position.

##### Solution :

At the extreme position, velocity of the pendulum is zero.$\\$ So there is no centrifugal force. $\\$ So T=mgcos$\theta_{\circ}$

17   17. A person stands on a spring balance at the equator. (a) By what fraction is the balance reading less than his true weight ? (b) If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case ?

##### Solution :

a)Net force on the spring balance. $\\$ R=mg -m$\omega^2$r $\\$ So the fraction less than the true weight (3mg) is $\\$ =$\frac{mg-(mg-m\omega^2r)}{mg}= \frac{\omega^2}{g}={[{\frac{2\pi}{24\times3600}}^2]}\times \frac{6400\times10^3}{10}=3.5\times 10^-3$ $\\$ b) When the balance reading is half the true weight, $\\$ $\frac{mg-(mg-m\omega^2r)}{mg}=1/2$ $\\$ $\omega^2r = g/2 \Rightarrow \omega= \sqrt{\frac{g}{2r}}=\sqrt{\frac{10}{2\times6400\times10^3}}rad/sec$ $\\$ $\therefore\hspace{0.2cm}Duration\hspace{0.2cm} of\hspace{0.2cm} the day is$ $\\$ $T = \frac{2\pi}{\omega} = 2\pi \times \sqrt{\frac{2\times6400\times10^3}{9.8}}sec=2\pi\sqrt{\frac{64\times10^6}{49}}sec= \frac{2\pi\times8000}{7\times3600}hr=2hr$ $\\$

18   18. A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up ?

##### Solution :

Given v=36km/hr =10m/s r=20m, $\mu=0.4$ $\\$ The road is banked with an angle , $\\$ $tan^-1[\frac{v^2}{rg}]=tan^-1[\frac{100}{20\times10}]=tan^-1 \frac{1}{2} or tan\theta=0.5$ $\\$ When the car travels at max speed such that it slips upwards, $\mu R1$ $\\$ acts downwards as shown in Fig.1 $\\$ So $R1 - mgcos\theta -\frac{mv1^2}{r}sin\theta=0 ..(i)$ $\\$ And $\mu R1 + mgsin\theta - \frac{mv1^2}{r}cos\theta = 0 ..(ii)$ $\\$ Solving the equations we get, $\\$ V1= $\sqrt{rg\frac{tan\theta- \mu }{1+\mu tan \theta}} =\sqrt{20\times10\times\frac{0.1}{1.2}} = 4.082 m/s = 14.7 km/hr$ $\\$ So the possible speeds lie between 14.7 km/hr and 54km/hr

19   19. A motorcycle has to move with a constant speed on an overbridge which is in the form of a circular arc of radius R and has a total length L. Suppose the motorcycle starts from the highest point. (a) What can its maximum velocity be for which the contact with the road is not broken at the highest point ? (b) If the motorcycle goes at speed 1/42 times the maximum found in part (a), where will it lose the contact with the road ? (c) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge ?

##### Solution :

So the possible speeds are between 15.7 km/hr and 54 km/hr $\\$ R= radius of the bridge $\\$ L=total length of the overbridge $\\$ a) At the highest point $\\$ mg=$\frac{mv^2}{R} \Rightarrow v^2 = Rg \Rightarrow v = \sqrt{Rg}$ $\\$ b) Given,$v =\frac{1}{\sqrt{2} }\sqrt{Rg}$ $\\$ suppose it loses contact at . So, at B, mgcos$\theta = \frac{mv^2}{R}$ $\\$ $\Rightarrow v^2=Rg_2cos\theta$ $\\$ $[\sqrt{\frac{Rv}{2}}]^2=Rgcos\theta \Rightarrow\frac{Rg}{2}=Rgcos\theta\Rightarrow cos\theta =\frac{1}{2} \Rightarrow60^{\circ}=\frac{\pi}{3}$ $\\$ $\theta=\frac{l}{r}\rightarrow l=r\theta=\frac{\pi R}{3}$ $\\$ So it will lose contact at distance $\frac{\pi R}{3}$ from the highest point $\\$ c) Let the uniform speed on the bridge be v. $\\$ The chance of losing contact is maximum at the end of the bridge for which a= $\frac{L}{2R}$ $\\$ So, $\frac{mv^2}{R}=mgcos\alpha \Rightarrow v=\sqrt{gRcos[\frac{l}{2R}] }$ $\\$

20   20. A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate —dv/dt = a. The friction coefficient between the road and the tyre is $\mu$. Find the speed at which the car will skid.

##### Solution :

Since the motion is non uniform, the acceleration has both radial and tangential components. $\\$ $a_r=\frac{V^2}{r}$ $\\$ $a_t=\frac{dv}{dt}=a$ $\\$ Resultant magnitude = $\sqrt{[\frac{v^2}{r}]^2+a^2}$ $\\$ Now $\mu N= \sqrt{[\frac{v^2}{r}]^2 + a^2} \Rightarrow \mu mg = m \sqrt{[\frac{v^2}{r}]^2+a^2}\Rightarrow\mu^2g^2=\frac{V^4}{r2}+a^2$ $\\$ $\Rightarrow v^4=(\mu^2g^2-a^2)r^2\Rightarrow v=[(\mu^2g^2-a^2)r^2]^{1/4}$

21   21. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is g. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip ? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration a, at what angular speed will the block slip ?

##### Solution :

22   21. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is g. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip ? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration a, at what angular speed will the block slip ?

##### Solution :

a) When the ruler makes uniform circular motion in the horizontal plane $\\$ $\mu=mg = m\omega_1^2L$ $\\$ $\omega_1=\sqrt{\frac{\mu g}{L}}$ $\\$ b) When the ruler makes uniformly accelerated circular motion $\\$ $\mu mg=\sqrt{({m\omega_2}^2L)^2 + (mL\alpha)^2}\Rightarrow \omega_2^4 + \alpha^2=\frac{\mu^2g^2}{L^2}\Rightarrow \omega_2=[(\frac{\mu g}{L})^2-\alpha^2]^{1/4}$

23   22. A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure (7-E1). Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track. (a) Find the normal contact force by the road on the cycle when it is at B and at D. (b) Find the force of friction exerted by the track on the tyres when the cycle is at B, C and D. (c) Find the normal force between the road and the cycle just before and just after the cycle crosses C. (d) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed ? Take g= 10 $\frac{m}{s^2}$

##### Solution :

Radius of the curves = 100m $\\$ Weight= 100Kg $\\$ Velocity = 18km/hr=5m/sec $\\$ a) at B mg$\frac{mv^2}{R}=N \Rightarrow N=(100\times10)-\frac{100\times25}{100}=1000-25=975N$ $\\$ At d, N= mg+$\frac{mv^2}{R}=1000+25=1025N$ $\\$ b)At B & D the cycle has no tendency to slide. So at B & D,frictional force is zero. $\\$ At 'C', mgsin$\theta = F \Rightarrow F = 1000\times \frac{1}{\sqrt2}=707N$ $\\$ c) (i) Before 'C' mgcos$\theta-N=\frac{mv^2}{R} \Rightarrow N= mgcos\theta-\frac{mv^2}{R}=707-25=683N$ $\\$ (ii) N-mgcos$\theta=\frac{mv^2}{R}+mgcos\theta=25+707=732N$ $\\$ d) To find out the minimum desired coeff. of friction,we have to consider a point just before C.( Where N is minimum) $\\$ Now, $\mu N= mgsin\theta \Rightarrow \mu\times682 = 707$ $\\$ So,$\mu=1.037$ $\\$

24   23. In a children's park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod (figure 7-E2). Let the mass of each kid be 15 kg, the distance between the points of the rod where the two kids hold it be 3.0 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids.

##### Solution :

d=3m$\Rightarrow R=1.5m$ $\\$ R=distance from the centre to one of the kids $\\$ N=20 rev per min=20/60=1/3 rev per sec $\\$ $\omega=2\pi r=2\pi /3$ $\\$ $m=15kg$ $\\$ $\therefore Frictional force F=mr\omega^2=15\times(1.5)\times\frac{(2\pi)^2}{9}=5 \times (0.5) \times 4\pi^2=10\pi^2$ $\\$ $\therefore$ Frictional force for one of the kids is $10\pi^2$ $\\$

25   24. A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle 0 with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is u. Find the range of the angular speed for which the block will not slip.

##### Solution :

If the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force acts downwards. Here, r$=Rsin\theta$ $\\$ From FBD-1 $\\$ R$_1-mgcos\theta-m\omega_1^2(Rsin\theta) sin\theta=0$ ..(i) [because r=$Rsin\theta$] $\\$ and $\mu R_1mgSin\theta-m\omega_1^2(Rsin\theta]$ $\\$ and $\mu R_1mgsin\theta-m\omega_1^2(Rsin\theta)cos\theta=0$ ..(ii) $\\$ Substituting the value of $R_1$ from Eq (i) in Eq(ii), it can be found out that $\\$ $\omega1=[\frac{g(sin\theta+\mu cos\theta}{Rsin\theta(cos\theta-\mu sin\theta)}]^{1/2}$ $\\$ Again, for minimum speed, the frictional force$\mu R_2$acts upward from FBD-2, it can be proved that $\\$ $\omega_2=[\frac{g(sin\theta-\mu cos\theta)}{Rsin\theta(cos\theta + \mu sin\theta)}]^{1/2}$ $\\$ $\therefore$ the range of speed is between $\omega_1$ and$\omega_2$ $\\$

26   25. A particle is projected with a speed u at an angle 0 with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle ? This radius is called the radius of curvature of the curve at the point.

##### Solution :

Particle is projected with speed 'u' at an angle $\theta$. At the highest point, the vertical component of velocity is '0'. $\\$ So, at the point, velocity =ucos$\theta$ $\\$ centripetal force = m$u^2cos^2(\frac{\theta}{r})$ $\\$ At the highest pt. $\\$ mg=$\frac{mv^2}{r} \Rightarrow r=\frac{u^2cos^2\theta}{g}$ $\\$

27   26. What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle 0/2 with the horizontal ?

##### Solution :

Let 'u' be the velocity at the pt where it makes an angle $\theta/2$ with the horizontal. The horizontal component remains unchanged $\\$ So, v cos$\theta/2=\omega cos\theta \Rightarrow v =\frac{ucos\theta}{cos\frac{\theta}{2}}$ ...(ii) $\\$ From the figure $\\$ mg$cos(\theta/2)=\frac{mv^2}{r} \Rightarrow r = \frac{V^2}{gcos(\theta/2)}$ $\\$ putting the value of 'v' from equn(i) $\\$ $r=\frac{u^2cos^2\theta}{gcos^3(\theta/2)}$ $\\$

28   27. A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is g. The block is given an initial speed v0. As a function of the speed v write (a) the normal force by the wall on the block, (b) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration ( $\frac{dv}{dt}$= $v\frac{dv}{dx}$) to obtain the speed of the block after one revolution.

##### Solution :

A block of mass 'm' moves on a horizontall circle against the wall of a cylindrical room of radius 'R'. $\\$ Friction coefficient between wall & the block is $\mu$ $\\$ a)Normal reaction by the wall on the block is $\frac{mv^2}{R}$ $\\$ b)$\therefore$ Frictional force by the wall =$\frac{\mu mv^2}{R}$ $\\$ c) $\frac{\mu mv^2}{R} = ma\Rightarrow a = \frac{-\mu v^2}{R}\Rightarrow ds=\frac{-R}{\mu}\frac{dv}{v}$ $\\$ d)Now, $\frac{dv}{dt}=v\frac{dv}{ds}=\frac{-\mu v^2}{R} \Rightarrow ds=\frac{-R}{\mu}\frac{dv}{v}$ $\\$ $\Rightarrow s =-R\mu ln V + C$ $\\$ At s=0,v=v$_{\circ}$ $\\$ $\therefore, c=\frac{R}{\mu}ln V_{\circ}$ $\\$ So ,s = $-\frac{R}{\mu}ln\frac{V}{V_{\circ}} \Rightarrow \frac{V}{V_{\circ}} = e^{-\mu s/R}$ $\\$ For one rotation $s=2\pi R$, so V=$V_{\circ}e^{-2_{pi \mu}}$ $\\$

29   28. A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity co in a circular path of radius R (figure 7-E3). A smooth groove AB of length L( «R) is made on the surface of the table. The groove makes an angle 0 with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move along AB. Find the time taken by the particle to reach the point B.

##### Solution :

The cabin rotates with angular velocity $\omega$ & radius R. $\\$ $\therefore$The particle experiences a force $mR\omega^2$ $\\$. The component of $mR\omega^2$ provides the required force to the particle to move along AB. $\\$ $\therefore mR\omega^2 cos\theta = ma \Rightarrow a=R\omega^2cos\theta$ $\\$ length of groove = L $\\$ L=ut+1/2$at^2\Rightarrow L=1/2R\omega^2 cos\theta t^2$ $\\$ $\Rightarrow t^2=\frac{2L}{R\omega^2cos\theta} \Rightarrow t = 1\sqrt{\frac{2L}{R\omega^2cos\theta}}$ $\\$

30   29. A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small wooden plate is kept on the seat with its plane perpendicular to the radius of the circular road (figure 7-E4). A small block of mass 100 g is kept on the seat which rests against the plate. The friction coefficient between the block and the plate is 1.1 = 0.58. (a) Find the normal contact force exerted by the plate on the block. (b) The plate is slowly turned so that the angle between the normal to the plate and the radius of the road slowly increases. Find the angle at which the block will just start sliding on the plate.

##### Solution :

v= Velocity of car= 36 km/hr =10m/s $\\$ r=Radius of circular path = 50m m= mass of small body = 100g =0.1kg $\\$ $\mu$= Friction coefficient between plates & body = 0.58 $\\$ a) The normal contact force exerted by the plate on the block $\\$ N=$\frac{mv^2}{r}=\frac{0.1\times 100}{50}=0.2N$ $\\$ b) The plate is turned so the angle between the normal to the plate & the radius of the road slowly increases $\\$ N=$\frac{mv^2}{r}cos\theta$ ..(ii) $\\$ $\mu N = \frac{mv^2}{r}sin\theta ..(ii)$ $\\$ Putting the value of N from (i) $\\$ $\mu \frac{mv^2}{r} cos\theta$ = $\frac{mv^2}{r}sin\theta \Rightarrow \mu=tan\theta \Rightarrow \theta = tan^-1\mu=tan^-1(0.58)=30^{\circ}$ $\\$

31   30. A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R . A smooth pulley of small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string going over the pulley. Initially the masses are held by a person with the strings along the outward radius and then the system is released from rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string.

##### Solution :

Let the bigger mass accelerate towards the right with acceleration 'a' $\\$ T-ma=$m\omega^2R=0$ ..(i) $\\$ T+2ma-2m$\omega^2R=0$ ..(ii) $\Rightarrow a =\frac{m\omega^2R}{3}$ $\\$ Substituting the value of a in Equation (i), we get T= 4/3m$\omega^2R.$ $\\$