Solve Rotational Collision Homework: 100 RPM to 50 RPM

[Soniteflash]

Homework Statement

A 2kg , 0.2 m diameter turntable rotates at 100 rpm on frictionless bearings. Two 0.5 kg block fall from above, hit the turntable simultaneously at opposite ends of the diameter, and stick. What is the turntable's angular velocity (in rpm) just after?

Homework Equations

L= I x ω
L_i = L_f
Moment of inertia for disk I = (1/2) MR²
Moment of inertia for point mass = MR²

The Attempt at a Solution

I converted 100 rpm to rotations per sec by dividing by 60 which gives me (5/3).
I thought about the problem and came up with the idea of conservation of angular momentum so
L_i = L_f.

L_i = I x ω= (1/2)MR² x ω. I am confused. The answer has to be in rpm but ω has the SI units of rad/s. Does unit consistency play a role here since I have ω on both sides of the equation.

I thought that before doing it with numbers, using only variables to make sure I understand what I am doing.

L_f= I_sys x ω_f So my question here is, does this reflect what is happening when the blocks stick to the plate? I reasoned that the blocks added moment of inertia to the system.

I continued and
I_sys = [((1/2)MR² x ω) + ( M_Block R² + M_Block R²)].

Is that correct?

Then I set them equal with L_i =L_f
and solved for angular velocity.

I continued and plugged in the given values and got a completely different answers. I feel like my approach is missing out on something.
The problem gave the answer which is 5.24 rad/sec or 50rpm.

 

[SammyS]

Homework Statement

A 2kg , 0.2 m diameter turntable rotates at 100 rpm on frictionless bearings. Two 0.5 kg block fall from above, hit the turntable simultaneously at opposite ends of the diameter, and stick. What is the turntable's angular velocity (in rpm) just after?

Homework Equations

L= I x ω
L_i = L_f
Moment of inertia for disk I = (1/2) MR²
Moment of inertia for point mass = MR²

The Attempt at a Solution

I converted 100 rpm to rotations per sec by dividing by 60 which gives me (5/3).
I thought about the problem and came up with the idea of conservation of angular momentum so
L_i = L_f.

L_i = I x ω= (1/2)MR² x ω. I am confused. The answer has to be in rpm but ω has the SI units of rad/s. Does unit consistency play a role here since I have ω on both sides of the equation.

I thought that before doing it with numbers, using only variables to make sure I understand what I am doing.

L_f= I_sys x ω_f So my question here is, does this reflect what is happening when the blocks stick to the plate? I reasoned that the blocks added moment of inertia to the system.

I continued and
I_sys = [((1/2)MR² x ω) + ( M_Block R² + M_Block R²)].

Is that correct?

Then I set them equal with L_i =L_f
and solved for angular velocity.

I continued and plugged in the given values and got a completely different answers. I feel like my approach is missing out on something.
The problem gave the answer which is 5.24 rad/sec or 50rpm.

There should be no ω in this expression for moment of inertia.

I_sys = [((1/2)MR² × ω) + ( M_Block R² + M_Block R²)].

Is that correct?

 

[gneill]

Homework Statement

A 2kg , 0.2 m diameter turntable rotates at 100 rpm on frictionless bearings. Two 0.5 kg block fall from above, hit the turntable simultaneously at opposite ends of the diameter, and stick. What is the turntable's angular velocity (in rpm) just after?

Homework Equations

L= I x ω
L_i = L_f
Moment of inertia for disk I = (1/2) MR²
Moment of inertia for point mass = MR²

The Attempt at a Solution

I converted 100 rpm to rotations per sec by dividing by 60 which gives me (5/3).

Careful, rotations per second is not the same as radians per second. How many radians in a rotation?

I thought about the problem and came up with the idea of conservation of angular momentum so
L_i = L_f.

L_i = I x ω= (1/2)MR² x ω. I am confused. The answer has to be in rpm but ω has the SI units of rad/s. Does unit consistency play a role here since I have ω on both sides of the equation.

Since rpm and radians per second are related by a proportionality constant you can get away with using "100 rpm" as the angular velocity in your equations. Note that the value you get for angular momentum will have "funny units" as a result, but that's resolved when you solve for the new angular velocity where the results will be in rpm.

I thought that before doing it with numbers, using only variables to make sure I understand what I am doing.

L_f= I_sys x ω_f So my question here is, does this reflect what is happening when the blocks stick to the plate? I reasoned that the blocks added moment of inertia to the system.

Yes, that's right.

I continued and
I_sys = [((1/2)MR² x ω) + ( M_Block R² + M_Block R²)].

Is that correct?

There's no "ω" in the moment of inertia calculation. ω gets involved when you are looking for the angular momentum.

Then I set them equal with L_i =L_f
and solved for angular velocity.

I continued and plugged in the given values and got a completely different answers. I feel like my approach is missing out on something.
The problem gave the answer which is 5.24 rad/sec or 50rpm.

Your problem may be due to the extraneous "ω" in your moment of inertia calculation, but it's hard to tell without seeing more of your intermediate steps.

 

[Soniteflash]

I see the mistake with omega!
For the conversion from rotations to radians. So i got the proportionality constant which gives me the number of rotations per second (5/3 rotations per sec). Then I can can multiply the PC by 2 PI and that gives me the radians?

 

[gneill]

I see the mistake with omega!
For the conversion from rotations to radians. So i got the proportionality constant which gives me the number of rotations per second (5/3 rotations per sec). Then I can can multiply the PC by 2 PI and that gives me the radians?

You can. So what's your value for radians per second?

 

[Soniteflash]

Oh my I forgot about this.
I apologize.
The value for radians per second: 10.47 rad/s