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Rotational motion

Question

A uniform rod of length L and mass M is lying on a frictionless horizontal plane and is pivoted at one of its ends as shown in figure. There is no friction at the pivot. An inelastic ball of mass m is fixed with the rod at a distance L/3 from O. A horizontal impulse J is given to the rod at a distance 2L/3 from O in a direction perpendicular to the rod. Assume that the ball remains in contact with the rod after the collision and impulse J acts for a small time interval Δt

Now answer the following questions: 

 

Moderate
Question

Find the resulting instantaneous angular velocity of the rod after the impulse.

Solution

Let the system starts with angular velocity ω. Angular velocity of the ball will also be ω as it remains struck to the rod. 

Velocity of the ball v=ωL3

For the rod, angular impulse = change in angular momentum:

J2L3JbL3=ML23ω            …(i)

For the ball, impulse = change in linear momentum

Jb=mv=L3               …(ii)

From Eqs. (i) and (ii), ω=6J(m+3M)L

and Jb=2mJ(m+3M)

Let impulse on the pivot be J1.

For the rod and ball system, total impulse = change in linear momentum : J+J1=Mvc+mv

Solve to get : J1=mJm+3M

Question

Find the impulse acted on the ball during the time interval Δt.

Solution

Let the system starts with angular velocity ω. Angular velocity of the ball will also be ω as it remains struck to the rod. 

Velocity of the ball v=ωL3

For the rod, angular impulse = change in angular momentum:

J2L3JbL3=ML23ω            …(i)

For the ball, impulse = change in linear momentum

Jb=mv=L3               …(ii)

From Eqs. (i) and (ii), ω=6J(m+3M)L

and Jb=2mJ(m+3M)

Let impulse on the pivot be J1.

For the rod and ball system, total impulse = change in linear momentum : J+J1=Mvc+mv

Solve to get : J1=mJm+3M

Question

Find the magnitude of the impulse applied by the pivot during the time interval Δt.

Solution

Let the system starts with angular velocity ω. Angular velocity of the ball will also be ω as it remains struck to the rod. 

Velocity of the ball v=ωL3

For the rod, angular impulse = change in angular momentum:

J2L3JbL3=ML23ω            …(i)

For the ball, impulse = change in linear momentum

Jb=mv=L3               …(ii)

From Eqs. (i) and (ii), ω=6J(m+3M)L

and Jb=2mJ(m+3M)

Let impulse on the pivot be J1.

For the rod and ball system, total impulse = change in linear momentum : J+J1=Mvc+mv

Solve to get : J1=mJm+3M



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