Q: A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle \theta to the horizontal, the maximum height attained by it equals 4R. The angle of projection \theta is then given by

\theta=\sin^{-1}\left(\dfrac{2gT^2}{\pi^2 R}\right)^{1/2}. Maximum height H_{max}=\dfrac{u^2\sin^2\theta}{2g}=4R (given). The speed in the circular path is u=\dfrac{2\pi R}{T}. Substituting, 4R=\dfrac{\left(\dfrac{2\pi R}{T}\right)^2\sin^2\theta}{2g}, which gives \sin\theta=\left(\dfrac{2gT^2}{\pi^2 R}\right)^{1/2}, so \theta=\sin^{-1}\left(\dfrac{2gT^2}{\pi^2 R}\right)^{1/2}.

Q: A particle moves in a circle of radius 5 cm with constant speed and time period 0.2\pi\,\text{s}. The acceleration of the particle is

5\,\text{m/s}^2. Given r=5\,\text{cm}=5\times10^{-2} m and T=0.2\pi s. The acceleration is a=r\omega^2=\dfrac{4\pi^2}{T^2}r=\dfrac{4\times\pi^2\times5\times10^{-2}}{(0.2\pi)^2}=5\,\text{m/s}^2.

Q: A particle moves along a circle of radius \left(\dfrac{20}{\pi}\right) m with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begin, the tangential acceleration is

40 m/s^2. a_T=r\alpha. Using \omega^2=\omega_0^2+2\alpha\theta with \omega_0=0, \omega=\dfrac{v}{r}=\dfrac{80}{20/\pi}=4\pi rad/s and \theta=2\times2\pi rad, we get \alpha=\dfrac{\omega^2}{2\theta}=\dfrac{(4\pi)^2}{2(4\pi)}=2\pi. Hence a_T=r\alpha=\dfrac{20}{\pi}\times2\pi=40 m/s^2.

Question 1

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle 	heta to the horizontal, the maximum height attained by it equals 4R. The angle of projection 	heta is then given by

Accepted Answer

heta=\sin^{-1}\left(\dfrac{2gT^2}{\pi^2 R}ight)^{1/2}. Maximum height H_{max}=\dfrac{u^2\sin^2	heta}{2g}=4R (given). The speed in the circular path is u=\dfrac{2\pi R}{T}. Substituting, 4R=\dfrac{\left(\dfrac{2\pi R}{T}ight)^2\sin^2	heta}{2g}, which gives \sin	heta=\left(\dfrac{2gT^2}{\pi^2 R}ight)^{1/2}, so 	heta=\sin^{-1}\left(\dfrac{2gT^2}{\pi^2 R}ight)^{1/2}.

Question 2

In the given figure, a=15\,	ext{m/s}^2 represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R=2.5 m at a given instant of time. The speed of the particle is

Accepted Answer

5.7 m/s. The centripetal acceleration is a_c=\dfrac{v^2}{R}. From the figure a_c=a\cos30^\circ=15\cos30^\circ, so \dfrac{v^2}{R}=15\cos30^\circ, giving v^2=R	imes15	imes\dfrac{\sqrt{3}}{2}=2.5	imes15	imes\dfrac{\sqrt{3}}{2}. Hence v=5.7 m/s.

Question 3

Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius \dfrac{r}{2} and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is

Accepted Answer

2. Equating the centripetal forces, (F_c)_{heavier}=(F_c)_{lighter}, so \dfrac{2m\,v^2}{r/2}=\dfrac{m\,(nv)^2}{r}. This gives n^2=4, hence n=2.

Question 4

A particle moves in a circle of radius 5 cm with constant speed and time period 0.2\pi\,	ext{s}. The acceleration of the particle is

Accepted Answer

5\,	ext{m/s}^2. Given r=5\,	ext{cm}=5	imes10^{-2} m and T=0.2\pi s. The acceleration is a=r\omega^2=\dfrac{4\pi^2}{T^2}r=\dfrac{4	imes\pi^2	imes5	imes10^{-2}}{(0.2\pi)^2}=5\,	ext{m/s}^2.

Question 5

A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is

Accepted Answer

0,\ 10 m/s. In one complete lap the displacement is zero, so average velocity =\dfrac{	ext{Displacement}}{	ext{Time}}=\dfrac{0}{t}=0. The distance travelled is 2\pi r, so average speed =\dfrac{2\pi r}{t}=\dfrac{2	imes3.14	imes100}{62.8}=10\,	ext{ms}^{-1}.

Question 6

A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 s, what is the magnitude and direction of acceleration of the stone?

Accepted Answer

\pi^2 ms^{-2} and direction along the radius towards the centre. Speed is constant, so it is uniform circular motion with radial acceleration a_r=r\omega^2=r\left(\dfrac{2\pi n}{t}ight)^2=\dfrac{1	imes4	imes\pi^2	imes(22)^2}{(44)^2}=\pi^2 m/s^2, directed along the radius towards the centre.

Question 7

The circular motion of a particle with constant speed is

Accepted Answer

periodic but not simple harmonic. In circular motion the particle repeats its motion after equal intervals of time, so it is periodic; but it is not simple harmonic as it does not execute to and fro motion about a fixed point.

Question 8

A particle moves along a circle of radius \left(\dfrac{20}{\pi}\right) m with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begin, the tangential acceleration is

Accepted Answer

40 m/s^2. a_T=r\alpha. Using \omega^2=\omega_0^2+2\alpha	heta with \omega_0=0, \omega=\dfrac{v}{r}=\dfrac{80}{20/\pi}=4\pi rad/s and 	heta=2	imes2\pi rad, we get \alpha=\dfrac{\omega^2}{2	heta}=\dfrac{(4\pi)^2}{2(4\pi)}=2\pi. Hence a_T=r\alpha=\dfrac{20}{\pi}	imes2\pi=40 m/s^2.

Question 9

P is the point of contact of a wheel and the ground. The radius of wheel is 1 m. The wheel rolls on the ground without slipping. The displacement of point P when wheel completes half rotation is

Accepted Answer

\sqrt{\pi^2+4} m. When the wheel completes half rotation without slipping, the horizontal displacement x=\pi R and the vertical displacement y=2R. So s=\sqrt{x^2+y^2}=\sqrt{(\pi R)^2+(2R)^2}=\sqrt{\pi^2 R^2+4R^2}. With R=1 m, s=\sqrt{\pi^2+4} m.

Question 10

A particle of mass M is revolving along a circle of radius R and another particle of mass m is revolving in a circle of radius r. If time periods of both particles are same, then the ratio of their angular velocities is

Accepted Answer

1. \omega=\dfrac{2\pi}{T}, so \omega\propto\dfrac{1}{T} and does not depend on mass or radius. \dfrac{\omega_1}{\omega_2}=\dfrac{T_2}{T_1}; since T_1=T_2, \dfrac{\omega_1}{\omega_2}=1.

Question 11

What is the linear velocity, if angular velocity vector \omega=3\hat{i}-4\hat{j}+\hat{k} and position vector r=5\hat{i}-6\hat{j}+6\hat{k}?

Accepted Answer

-18\hat{i}-13\hat{j}+2\hat{k}. \mathbf{v}=\boldsymbol{\omega}	imes\mathbf{r}=(3\hat{i}-4\hat{j}+\hat{k})	imes(5\hat{i}-6\hat{j}+6\hat{k})=\hat{i}(-24+6)-\hat{j}(18-5)+\hat{k}(-18+20)=-18\hat{i}-13\hat{j}+2\hat{k}.

Question 12

A body is whirled in a horizontal circle of radius 20 cm. It has an angular velocity of 10 rad/s. What is its linear velocity at any point on circular path?

Accepted Answer

2 m/s. Linear speed v=r\omega. Here r=20 cm =0.20 m and \omega=10 rad/s, so v=0.20	imes10=2 m/s.

Question 13

When milk is churned, cream gets separated due to

Accepted Answer

centrifugal force. By the concept of centrifugal force the cream is separated from milk. A mass m of milk at distance r from the axis requires a centripetal force mr\omega^2; lighter cream particles of mass m'<m require less, so under the net force (m-m')r\omega^2 the cream moves towards the axis of rotation, and on stopping the centrifuge the cream is found at the top and the milk at the bottom.

Question 14

An electric fan has blades of length 30 cm measured from the axis of rotation. If the fan is rotating at 120 rev/min, the acceleration of a point on the tip of the blade is

Accepted Answer

47.4 ms^{-2}. Centripetal acceleration a_c=\dfrac{v^2}{r}=\dfrac{r^2\omega^2}{r}=r\omega^2 (as v=r\omega), with \omega=2\pi
u. Here r=30 cm =0.30 m and 
u=120 rev/min =\dfrac{120}{60}=2 rev/s. So a_c=r(2\pi
u)^2=0.30	imes4	imes3.14	imes3.14	imes2	imes2=47.4 ms^{-2}.

Uniform Circular Motion — NEET Physics PYQ MCQs with Solutions

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