Dynamics of Circular Motion — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Dynamics of Circular Motion MCQs with step-by-step solutions (9 questions). Part of Laws of Motion. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Dynamics of Circular Motion · medium
A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be ($g=10$ m/s$^2$)
A. $\dfrac{10}{2\pi}$ rad/s
B. 10 rad/s ✓ Correct
C. $10\pi$ rad/s
D. $\sqrt{10}$ rad/s
Solution: The block stays stationary when the limiting friction is at least the weight: $f_l\geq mg$, with $f_l=\mu N$ and $N=mr\omega^2$. Thus $\mu mr\omega^2\geq mg\Rightarrow\omega\geq\sqrt{\dfrac{g}{r\mu}}$. So $\omega_{min}=\sqrt{\dfrac{g}{r\mu}}=\sqrt{\dfrac{10}{1\times0.1}}=10$ rad/s.
Q2 — Dynamics of Circular Motion · medium
One end of the string of length $l$ is connected to a particle of mass $m$ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed $v$, the net force on the particle (directed towards center) will be ($T$ represents the tension in the string)
A. $T$ ✓ Correct
B. $T+\dfrac{mv^2}{l}$
C. $T-\dfrac{mv^2}{l}$
D. Zero
Solution: In uniform circular motion the net force equals the centripetal force, which is provided entirely by the tension: $\dfrac{mv^2}{l}=T$. Hence the net force directed towards the centre is $T$.
Q3 — Dynamics of Circular Motion · medium
A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s$^{-2}$. Its net acceleration in ms$^{-2}$ at the end of 2.0 s is approximately
A. 7.0
B. 6.0
C. 3.0
D. 8.0 ✓ Correct
Solution: Angular speed $\omega=\alpha t=2\times2=4$ rad s$^{-1}$. Centripetal acceleration $a_c=\omega^2 r=4^2\times0.5=8$ m/s$^2$; tangential acceleration $a_t=\alpha r=2\times0.5=1$ m/s$^2$. Net acceleration $a=\sqrt{a_c^2+a_t^2}=\sqrt{8^2+1^2}=\sqrt{65}\approx8$ m/s$^2$.
Q4 — Dynamics of Circular Motion · medium
A car is negotiating a curved road of radius $R$. The road is banked at angle $\theta$. The coefficient of friction between the tyres of the car and the road is $\mu_s$. The maximum safe velocity on this road is
A. $\sqrt{gR\left(\dfrac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}\right)}$ ✓ Correct
B. $\sqrt{\dfrac{g}{R}\left(\dfrac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}\right)}$
C. $\sqrt{\dfrac{g}{R^2}\left(\dfrac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}\right)}$
D. $\sqrt{gR^2\left(\dfrac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}\right)}$
Solution: Vertical equilibrium: $N\cos\theta=mg+f_s\sin\theta$; horizontal: $N\sin\theta+f_s\cos\theta=\dfrac{mv^2}{R}$. With $f_s=\mu_s N$, dividing gives $\dfrac{v^2}{Rg}=\dfrac{\sin\theta+\mu_s\cos\theta}{\cos\theta-\mu_s\sin\theta}$, so $v=\sqrt{gR\left(\dfrac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}\right)}$.
Q5 — Dynamics of Circular Motion · medium
A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is $45^\circ$, the speed of the car is
A. 20 ms$^{-1}$
B. 30 ms$^{-1}$ ✓ Correct
C. 5 ms$^{-1}$
D. 10 ms$^{-1}$
Solution: For a frictionless banked road $\tan\theta=\dfrac{v^2}{rg}$. With $\theta=45^\circ$, $r=90$ m and $g=10$ m/s$^2$: $v=\sqrt{rg\tan45^\circ}=\sqrt{90\times10\times1}=30$ m/s.
Q6 — Dynamics of Circular Motion · medium
A gramophone record is revolving with an angular velocity $\omega$. A coin is placed at a distance $r$ from the centre of the record. The static coefficient of friction is $\mu$. The coin will revolve with the record if
A. $r=\mu g\omega^2$
B. $r<\dfrac{\omega^2}{\mu g}$
C. $r\leq\dfrac{\mu g}{\omega^2}$ ✓ Correct
D. $r\geq\dfrac{\mu g}{\omega^2}$
Solution: The frictional force $\mu mg$ must supply the centripetal force: $F_{friction}\geq F_{centripetal}$, i.e. $\mu mg\geq m\omega^2 r$, so $\dfrac{\mu g}{r}\geq\omega^2$, giving $r\leq\dfrac{\mu g}{\omega^2}$.
Q7 — Dynamics of Circular Motion · medium
A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The string will break if the tension is more than 25 N. What is the maximum speed with which the ball can be moved?
A. 14 m/s ✓ Correct
B. 3 m/s
C. 3.92 m/s
D. 5 m/s
Solution: For horizontal circular motion, tension = centripetal force: $T_{max}=\dfrac{Mv_{max}^2}{R}$, so $v_{max}=\sqrt{\dfrac{T_{max}\cdot R}{M}}=\sqrt{\dfrac{25\times1.96}{0.25}}=\sqrt{196}=14$ m/s.
Q8 — Dynamics of Circular Motion · medium
What will be the maximum speed of a car on a road turn of radius 30 m, if the coefficient of friction between the tyres and the road is 0.4? (Take $g=9.8$ m/s$^2$)
A. 10.84 m/s ✓ Correct
B. 9.84 m/s
C. 8.84 m/s
D. 6.84 m/s
Solution: The centripetal force is provided by friction: $\dfrac{mv^2}{r}\leq\mu mg\Rightarrow v\leq\sqrt{\mu rg}$. So $v_{max}=\sqrt{\mu rg}=\sqrt{0.4\times30\times9.8}=10.84$ m/s.
Q9 — Dynamics of Circular Motion · medium
Two racing cars of masses $m$ and $4m$ are moving in circles of radii $r$ and $2r$ respectively. If their speeds are such that each makes a complete circle in the same time, then the ratio of the angular speeds of the first to the second car is
A. 8 : 1
B. 4 : 1
C. 2 : 1
D. 1 : 1 ✓ Correct
Solution: Both cars complete the circle in the same time and $\omega=\dfrac{2\pi}{t}$, so their angular speeds are equal. Hence the ratio of the angular speeds is $1:1$.