Answer

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**Hint:**The change in angular momentum with respect to time on a body is called the torque acting on that body. Also change in velocity is the linear momentum the body possesses. Also, angular momentum is affected by linear momentum for a point particle.

Formula used: We will be using the formulas concerning conservation of force according to the free body diagram. Also we will be using, $ \tau = rF\sin \theta $ where $ r $ is the radius of the surface on which force is acting, $ F $ is the force that acts on the surface, and $ \theta $ is the angle between the vector that connects the $ F $ and $ r $ vector. Also we will be using, $ L = rmv\sin \phi $ where $ L $ is the angular momentum of a point, $ r $ is the radius upon which it acts, $ m $ is the mass of the object which causes the momentum, $ v $ is the velocity at which the object moves, and $ \phi $ is the angle between the radius vector and velocity vector. We might also use the formula $ \tau = \dfrac{{dL}}{{dt}} $ where $ \tau $ is the torque, $ L $ is the angular momentum, and $ t $ is the time.

**Complete Step by step answer**

Observing the free body diagram for the question, we can find which of the forces are conserved in the system.

As you can see, the vertical cylinder is mounted in a horizontal plane (not shown in fig). The cylinder is attached to a particle via thread AB. Thus, it is touching the cylinder at B. Initially the particle is at rest at the position A.

At this position, the particle will experience 3 major forces. The gravitational force that pulls the particle downwards $ Fg = mg $ where $ Fg $ is the gravitational force, $ m $ is the mass of the object, and $ g $ is the acceleration due to gravity. Also, a normal force $ N $ acts opposite to gravity. Obviously, there is tension $ T $ created on the thread AB.

$ N = mg $ Thus, these forces are conserved. While tension $ T $ is left unbalanced.

Also note that the particle at A is in motion along $ {v_0} $ , thus producing a torque $ \tau $ at B and also at O (the centre of the circular surface below).

[ Since , $ \tau = rmv0\sin \theta $ and none of the variables are zero, there is a torque $ \tau $ ]

So now we know for sure that the torque at points B and O, $ \tau \ne 0 $ .

Also, we know that Torque is also defined as the rate of change of angular momentum.

$ \Rightarrow \tau = \dfrac{{dL}}{{dt}} $

Thus if $ \tau \ne 0 $ , the $ \dfrac{{dL}}{{dt}} \ne 0 $ .(Change in angular momentum $ \ne $ 0)

$ \Rightarrow $ The angular momentum at B and O are not constant. This makes the options A and B false.

Now, we know that the particle begins its motion at A with velocity $ v0 $ . However, due to the tension on thread AB, the direction of velocity keeps changing. Thus, implying there is change in momentum due to change in velocity.

$ \Rightarrow $ The momentum is not constant. Thus, making option C false.

Now, we can also observe that the Tension $ T $ has not caused any work to be done, nor the system has dissipated any kind of energy. So, the kinetic energy of the system is constant.

$ \Rightarrow $ The kinetic energy of the system is constant. Thus, making option D true.

The options that are not true among the ones provided are A, B, C.

**Note**

The torque $ \tau = rmv\sin \theta $ is not necessary to find the existence of torque in the above problem. We know if a force causes a motion around a specific axis then that force is called torque. However, it was only used to introduce the relationship between angular momentum for a point object with linear momentum of the same object.

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