Kinematics of Circular Motion

A particle of mass $m$ moves along a circle of radius $R$ with a normal acceleration varying with time as, $a_{\mathrm{r}}=a{t}^2$, where $a$ is a constant. Find the time dependence of the power $\left(P\right)$, developed by all the forces acting on the particle, and the mean value of this power  averaged over a time $t$ after the beginning of the motion.

A particle $A$ moves along a circle of radius, $R=50 \mathrm{cm}$so that its radius vector $r$, relative to the point $O$ (figure), rotates with the constant angular velocity, $\omega =0.40 \mathrm{rad} {\mathrm{s}}^{-1}$. Then, the modulus of the velocity $\left(v\right)$ of the particle and the modulus of its total acceleration $\left(a\right)$ will be,

A spotlight $S$ rotates in a horizontal plane with a constant angular velocity of $0.1 \mathrm{rad} {\mathrm{s}}^{-1}$. The spot of the light, $P$ moves along the wall at a distance $3 \mathrm{m}$. What is the velocity of the spot $P$ when $\text{θ}=45°$?

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

A stone tied to one end of string, $80 \mathrm{cm}$ long, is whirled in a horizontal circle at a constant speed. If stone makes $25$revolutions in $14 \mathrm{s}$, the magnitude of the acceleration of stone is (approximately)

A stone tied to the end of a string of $1 \mathrm{m}$ long is whirled in a horizontal circle with a constant speed. If the stone makes $22$ revolutions in $44 \mathrm{s}$, what is the acceleration of the stone?

A particle moves along an arc of a circle of radius $R$. Its velocity depends on the distance covered $s$ as $v=a\sqrt{s}$, where, $a$ is a constant. Then the angle between the vector of total acceleration and the vector of velocity as a function of $s$ will be?

The magnitude of displacement of a particle moving in a circle of the radius $a$ with constant angular speed $\omega$ varies with time $t$ as

For a particle moving with speed $v$ in a uniform circular motion, the acceleration at a point $P$$\left(R, \mathrm{\theta }\right)$ in the first quadrant on the circle of radius $R$ is $($here, $\mathrm{\theta }$ is measured from the $x$-axis$)$

The figure shows the total acceleration and velocity of a particle moving clockwise in a circle of radius $2.5 \mathrm{m}$ at a given instant of time. At this instant, find $\left(\mathrm{i}\right)$ its radial acceleration,$\left(\mathrm{ii}\right)$ the speed of the particle and $\left(\mathrm{iii}\right)$ its tangential acceleration.

The square of the angular velocity $\omega \left(\mathrm{in} \mathrm{rad} {\mathrm{s}}^{-1}\right)$ of a certain wheel increases linearly with the angular displacement during $100$revolutions of the wheel's motion as shown. Compute the time $t \left(\mathrm{in} \mathrm{rad}\right)$ for given $100$ revolutions.

The shaft of an electric motor starts from rest and on the application of torque, it gains an angular acceleration given by $\alpha =3t-t^2\left(\mathrm{in} \mathrm{rad} {\mathrm{s}}^{-2}\right)$ during the first $2 \mathrm{s}$, after which it becomes equal to zero. The angular velocity after $6 \mathrm{s}$ will be?

A particle is revolving in a circle of the radius $R$ with initial speed $v$. It starts retarding with constant retardation $\frac{v^2}{4\mathrm{\pi }R}$. The number of revolutions it makes before coming to rest is

A particle performs circular motion of radius $1 \mathrm{m}$ from rest. The tangential acceleration of the particle at any time $t \left(\mathrm{in} \mathrm{s}\right)$ is given by $a_{\mathrm{t}}=t \left(\mathrm{in} \mathrm{m} {\mathrm{s}}^{-2}\right)$. The radial acceleration of the particle at $t=2 \mathrm{s}$ is

A particle begins to move with a tangential acceleration of constant magnitude $0.6 \mathrm{m} {\mathrm{s}}^{-2}$ in a circular path. If it slips when its total acceleration becomes $1 \mathrm{m} {\mathrm{s}}^{-2}$, then the angle through which it would have turned before it started to slip is

A point $P$ moves in a counter-clockwise direction on a circular path as shown in the figure. The movement of $P$ is such that it sweeps out a length $s=t^3+5$, where $s$ is in $\mathrm{m}$ and $t$ is in$\mathrm{seconds}$. The radius of the path is $20 \mathrm{m}$. The acceleration of $P$ when $t=2 \mathrm{s}$ is nearly

Which of the following statements is false for a particle moving in a circle with a constant angular speed?

The angular displacement of any particle is given as $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$, where ${\omega }_0$ and $\alpha$ are constants. If ${\omega }_0=1 \mathrm{rad} {\mathrm{s}}^{-1}$, $\alpha =1.5 \mathrm{rad} {\mathrm{s}}^{-2}$ then, its angular velocity at $t=2 \mathrm{s}$ $($ in $\mathrm{rad} {\mathrm{s}}^{-1}$) will be

A particle is kept fixed on a uniformly rotating turn-table. As seen from the ground, the particle goes in a circle and its speed and acceleration are $10 \mathrm{cm} {\mathrm{s}}^{-1}$ and $10 \mathrm{cm} {\mathrm{s}}^{-2}$ respectively. The particle is now shifted to a new position to make the radius double of the original value. The new values of the speed and acceleration will respectively, be

A grind-stone starts revolving from rest and its angular acceleration is $4.0 \mathrm{rad} {\mathrm{s}}^{-2}$ (uniform). Then after $4 \mathrm{s}$, what is its angular displacement and angular velocity, respectively?