Rotational kinematics of a spherical rock upon collision

[Nerrad]

Homework Statement

A small spherical rock of mass collides in space with a large spherical rock of mass as indicated in the diagram. After the collision the rocks stick together to form a single spherical object.
https://postimg.org/image/fltmg3bj5/
(New here so I've no clue how to upload photo) https://postimg.org/image/fltmg3bj5/

Note: I only need help with Question d)

Consider the frame of reference in which the large rock is initially stationary and the small rock is moving with velocity as shown. You may assume that |v| ≪ c.

a) Explain in words what happens when the two rocks collide with the condition a = 0 and also what happens in the case that is not equal to zero.

b) If a= 0.0m , R= 1 m, m= 1kg , M = 10000 and |v| = 100m /s , calculate the velocity of the object after the collision.

c) If a= 0.5 m calculate the velocity of the object after the collision and comment on your result.

d) For c) calculate the total angular momentum about of the system and hence the rotation rate of the object after the collision. You may assume that the change in radius is negligible after the collision and use the approximation m≪M . [Hint. The moment of inertia of a sphere is 2/5 MR^2 .]

Homework Equations

The Attempt at a Solution

From the previous questions I found out that the linear velocity of the rock to be 0.01m/s. So I used v=Rw to find the angular velocity and then use (Angular Momentum equation) L = Iw and using I as (2/5)MR^2. I got 40kgm^2/s but the answer says it is 50 kgm/s^2 and the angular velocity is 0.0125rad/s. What have I done wrong?

 

[Doc Al]

Hint: The angular momentum of the system equals the angular momentum of the small rock before the collision.

 

[haruspex]

I used v=Rw to find the angular velocity and then use (Angular Momentum equation) L = Iw and using I as (2/5)MR^2. I got

We can rarely pinpoint errors without seeing the working.

 

[Nerrad]

We can rarely pinpoint errors without seeing the working.

So the linear velocity after the conservation of momentum came out to be 0.01m/s. Using w = v/R where R is the radius of the big rock after collision, w = 0.01/1 = 0.01rad/s. L = Iw = (2/5) MR^2 * w = ((2/5)*(10000kg)*(1m)^2)*0.01m/s = 40kgm^2/s. However the answer comes out to be 50kgm^2/s and w comes out to be 0.0125rad/s

 

[PeroK]

So the linear velocity after the conservation of momentum came out to be 0.01m/s. Using w = v/R where R is the radius of the big rock after collision, w = 0.01/1 = 0.01rad/s. L = Iw = (2/5) MR^2 * w = ((2/5)*(10000kg)*(1m)^2)*0.01m/s = 40kgm^2/s. However the answer comes out to be 50kgm^2/s and w comes out to be 0.0125rad/s

$w = v/R$ is not appropriate here, as $v$ is the linear speed of the rock.

 

[haruspex]

So the linear velocity after the conservation of momentum came out to be 0.01m/s. Using w = v/R where R is the radius of the big rock after collision, w = 0.01/1 = 0.01rad/s. L = Iw = (2/5) MR^2 * w = ((2/5)*(10000kg)*(1m)^2)*0.01m/s = 40kgm^2/s. However the answer comes out to be 50kgm^2/s and w comes out to be 0.0125rad/s

To elaborate on PeroK's reply, the motion of the large rock after collision consists of a linear motion of the mass centre, as you calculated, plus a rotation about the mass centre.

 

[Nerrad]

$w = v/R$ is not appropriate here, as $v$ is the linear speed of the rock.

So I get it now that the v in v=rw applies to tangential velocity and you're saying that the velocity obtained after conservation of momentum is the linear velocity of the rock as a whole so it v =rw wouldn't apply? What is the relationship between linear velocity and tangential velocity?Or does the linear velocity has nothing to do with the rotational motion after?

 

[Nerrad]

Hint: The angular momentum of the system equals the angular momentum of the small rock before the collision.

I don't see how you could calculate the angular momentum of the small rock before the collision since L=Iw= (2/5)(MR^2)*(v/R) = (2/5)(MRv)=(2/5)(1)(1)(100)=40??

 

[PeroK]

So I get it now that the v in v=rw applies to tangential velocity and you're saying that the velocity obtained after conservation of momentum is the linear velocity of the rock as a whole so it v =rw wouldn't apply? What is the relationship between linear velocity and tangential velocity?Or does the linear velocity has nothing to do with the rotational motion after?

The linear velocity of the CoM is independent of the angular velocity about the centre of mass. In part a) of thsi question you have a linear velocity of $v$ but no rotation.

 

[Doc Al]

I don't see how you could calculate the angular momentum of the small rock before the collision since L=Iw= (2/5)(MR^2)*(v/R) = (2/5)(MRv)=(2/5)(1)(1)(100)=40??

The angular momentum of the small rock has nothing to do with the rotation of the big rock (at least not before they collide!). You need to learn how to find the angular momentum of a particle.

 

[PeroK]

I don't see how you could calculate the angular momentum of the small rock before the collision since L=Iw= (2/5)(MR^2)*(v/R) = (2/5)(MRv)=(2/5)(1)(1)(100)=40??

In this case we have angular momentum of the system about any given point. Any object in linear motion (in this case the small rock) has non-zero angular momentum about any point not on its line of motion.

 

[Nerrad]

The linear velocity of the CoM is independent of the angular velocity about the centre of mass. In part a) of thsi question you have a linear velocity of $v$ but no rotation.

I understood that when in part a) there's no rotation because a = 0. I noted that my question is Qd) which asked for the angular velocity and angular momentum when a = 0.5, so surely there's rotation?

 

[PeroK]

I understood that when in part a) there's no rotation because a = 0. I noted that my question is Qd) which asked for the angular velocity and angular momentum when a = 0.5, so surely there's rotation?

Yes, but where does the rotation, hence angular momentum about the CoM come from? By conservation of angular momentum, there must be angular momentum (in the system) before the collision. That comes from the linear motion of the rock.

Angular momentum is not only about rotation and orbital motion. Linear momentum is also a special case of angular momentum.

 

[Doc Al]

Read this: Angular Momentum of a Particle

 

[Nerrad]

Yes, but where does the rotation, hence angular momentum about the CoM come from? By conservation of angular momentum, there must be angular momentum (in the system) before the collision. That comes from the linear motion of the rock.

Angular momentum is not only about rotation and orbital motion. Linear momentum is also a special case of angular momentum.

So the angular momentum of the small rock about point O before the collision is its mass m * distance from point O * its angular velcoity about point O. But the angular velocity is constantly changing?

 

[Doc Al]

So the angular momentum of the small rock about point O before the collision is its mass m * distance from point O * its angular velcoity about point O.

No, not its angular velocity. (Read the link I gave.)

 

[Nerrad]

No, not its angular velocity. (Read the link I gave.)

So the angular momentum L = mvr which taking r as a=0.5, gives a result of L=50 which is apparently the right answer. But why would r be a if its rotating about point O?

 

[Doc Al]

So the angular momentum L = mvr which taking r as a=0.5, gives a result of L=50 which is apparently the right answer. But why would r be a ?

Because what you really want is the angular momentum of the system about its center of mass. Which, for all practical purposes, is the center of the big rock.

 

[Nerrad]

Because what you really want is the angular momentum of the system about its center of mass. Which, for all practical purposes, is the center of the big rock.

Then the distance to the centre of mass of the big is constantly changing as the small rock approaches the CoM?

 

[PeroK]

So the angular momentum L = mvr which taking r as a=0.5, gives a result of L=50 which is apparently the right answer. But why would r be a ?

Angular momentum is a vector:

$\vec{L} = \vec{r} \times m\vec{v}$

For motion in a plane this can be simplified to an equation involving signed scalars:

$L = mv r \sin \theta = mva$

 

[Doc Al]

Then the distance to the centre of mass of the big is constantly changing as the small rock approaches the CoM?

But what matters is the perpendicular distance from the origin to the line of velocity of the small rock. That equals "a".

 

[Nerrad]

But what matters is the perpendicular distance from the origin to the line of velocity of the small rock. That equals "a".

Thanks guys I appreciate this alot. :)

 

[akshay86]

by momentum conservation (M+m)V=mv
V=mv/(M+m)
force on large rock,F=change in momentum by time=m(v-V)/t=mv(1-1/(M+m))/t
torque on it,T=RFsinx=RFa/R=Fa
ang.momentum,L=Tt=Fat=mv(1-1/(M+m))a