Circular Motion Practice Problems with Answers Physics Class 11

Ans. A particle can be accelerated if its velocitychanges. A particle having uniform circular moton has constant speed but its dirction of motion changes continuously. Due to this, there is change in velocity and hence the particle is accelerated. Is it possible for a body to have constant speed but variable velocity? If so, give example.

Ans. Hes. When a body moves in a circle with constant speed, its velocity is variable. This because velocity is a vector quantity whose magnitude is speed but its direction changes continuously due to the change of the direction of motion.

Ans. No. Velocity is a vector quantity and speed is its magnitude. A vector quantity changes if either its magnitude or direction changes. Since the magnitude of velocity (i.e., speed) changes and hence the body can not have a constant velocity.

Ans. Circular motion is possible at constant speed because in circular motion, the magnitude of the velocity i.e., speed remains constant while the direction of motion changes continuously.

Ans. No, since acceleration is always towards the centre.

Ans. Kinetic energy and speed.

Ans. The two particles should be moving in circular path of radius d / 2 with same speed. Since tangent at any point gives the direction of velocity, they will always be opposite. They will turn to their initial position after completing one circular path.

Ans. When the car turns round a curve, the passengers sitting in the car experience an outward force, i.e. centrifugal force due to the absence of the necessary centripetal force.

Ans. Yes, when a particle moves in a circles with constant speed, its velocity is variable because the direction of motion is continuously changing. accelerating.

Ans. It is a device used for separating mixtures of substances with different densities. It works on the principle that the centrifugal force, $F=m r w^{2}$ acting on a body in circular motion is, (i) Directly proportional to mass ( $m$ ) of the body. (ii) Directly proportional to the distance of the body from its axis of rotation. The heavier particles experience large amounts of centrifugal force, i.e., they will be push outward, and get separated from the lighter particles.

Ans. For revolving round the earth, the moon provides itself the necessary centripetal force at the expense of its weight. We know that a body falls under the effect of its weight. Since the weight of the moon gets used in providing it the necessary centripetal force, it revolves round the earth, without falling towards it.

Ans. The direction of the centripetal force does not depend, whether the body is moving in clockwise or anticlockwise direction. It is always directed along the radius and towards the centre of the circle.

Ans. While going along a curve, the cyclist leans to one side, so as to provide himself the necessary centripetal force. He leans in the inward direction i.e., towards the centre of the circular curve.

Ans. $l=1 m,$ applying conservation of energy, we have $m g(1-\cos 60)=\frac{1}{2} m v_{B}^{2}$ $V_{B}=2 g\left(1-\frac{1}{2}\right)=g$ Centripetal acceleration $=g \sin 60^{\circ}=\frac{\sqrt{3}}{2} g$ Velocity difference $=v_{B}-v_{A}=10-4 m s^{-1}$

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Ans. since velocity changes with time, the motion in circular path involves tangential acceleration. So it is a non-uniform circular motion. Net acceleration $=a_{n} \sqrt{a_{r}^{2}+a_{t}^{2}}=\sqrt{\left(K^{2} r t^{2}\right)^{2}+(K r)^{2}}=K r \sqrt{1+K^{2} t^{4}}$

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Ans. On a plane road, the centripetal force required to negotiate the curve is provided by the force of friction between the tyres of the vehicle and the surface of the road. Thus, the force of friction depends upon the nature of the surfaces. The force of friction may provide the necessary centripetal force if the speed of the vehicle is small enough. On the other hand, if the speed of the vehicle at the curve is not small enough, then theforce of friction will be unable to provide the necessary centripetal force to the vehicle and it may skid away. Therefore, to provide the necessary centripetal force to the vehicle at higher speeds, roads are banked.

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Ans. Given that and The centripetal force required for negotiating the curve on the unbanked road is provided by the lateral thrust by the outer rail on the flanges of the wheels. According to Newton's third law of motion the train exerts an equal an opposite force on the outer rail. Hence, the outer rail will wear out faster than the inner rail.Further, given that,Speed of train, Radius of the bend, . If is the angle of banking required to prevent wearing out of the rails,     $\theta=37^{\circ} 25^{\prime}$

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Ans. The various forces acting on $M$ and $m$ are $T=M g$ $T \cos \theta=m g$ $\begin{array}{l}{T \sin \theta=\frac{m v^{2}}{r}=m r \omega^{2}} \\ {\text { Where } r \text { is the radius of the circular path and } \omega} \\ {\text { is the angular velocity. }} \\ {r=l \sin \theta} & {\text { [From(iii)] }}\end{array}$  

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Ans. $v=7.5 \mathrm{ms}^{-1}, r=80 \mathrm{m}$ centripetal acceleration is, When the cyclist applies brakes at $P$ of the circular turn, the cyclist applies brakes at $P$ of the circular Acceleration will act opposite to the velocity i.e., $a_{t}=0.5 \mathrm{ms}^{-2}$ $a=\sqrt{a_{c}^{2}+a_{t}^{2}}=\sqrt{(0.7)^{2}+(0.5)^{2}}=0.86 \mathrm{ms}^{-2}$ Let $\theta$ be the angle made by net acceleration with the velocity of the cyclist, then $\tan \theta=\frac{a_{c}}{a_{t}}=\frac{0.7}{0.5}=1.4$

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Ans. The car is acted upon in the Vertical  direction by two forces: its weight $M g$ and the normal reaction $R$. (a) When the car runs over a horizontal bridge, the normal reaction, say $R_{1}$ is just equal and opposite to its weight $M g$ i.e., $R_{1}=M g$ Thus, force on the bridge is equal to the weight of the car. (b) When the car runs over a convex bridge, it requires a centripetal force vertically downwards. If $R_{2}$ is normal reaction, then Thus, force on the bridge is less then the weight of the car.

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Ans. When the bob is released from the horizontal position. A, it will reach the equilibrium position by moving along circular path as shown figure. At point , the bob of the pendulum is acted upon by two forces: its weight m g and tension in the string. The resultant of these forces provides the necessary centripetal force i.e., Kinetic energy at of the bob point $O=$ Potential energy at point $A$ $\frac{1}{2} m v^{2}=m g l \Rightarrow v=\sqrt{2 g l}$ Substituting for $v$ in equation (i), we have The string should have minimum strength to stand a tension three times the weight of the bob.

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