How Does a Frog's Jump Affect Disk Rotation Time?

[Chase11]

Homework Statement

a) A 1.5 kg frog sits at rest on top of a solid disk. The disk's mass is 4 kg, its radius is 1.25 m, and it rotates on a frictionless axle. The frog jumps from the disk at 3.7 m/s at an angle of 50°. Find the angular momentum of the frog about the axle as it leaves the disk.
b)How much time will it take the disk in this problem to rotate one complete revolution after the frog jumps?

Homework Equations

a) ∠=mvrsinσ

b)v=d/t or 2∏r/v=t

The Attempt at a Solution

I was able to find the angular momentum of the frog just fine. ∠=-4.46kgm^2/s. My problem is finding the time it takes the disk to complete one revolution after the frog has jumped. I used t=d/v or 2∏r/v=2∏(1.25m)/3.7m/s. The answer is supposed to be 4.40 seconds and I am getting something different. Am I setting up this problem incorrectly?

 

[collinsmark]

Hello Chase11,

Welcome to Physics Forums!

I used t=d/v or 2∏r/v=2∏(1.25m)/3.7m/s. The answer is supposed to be 4.40 seconds and I am getting something different. Am I setting up this problem incorrectly?

But that method won't work. You don't know what the velocity of the edge of the disk is (you were assuming it's 3.7 m/s, but it's not. That's the frog's velocity, not the disk's) [Edit: by that I mean you don't know what the velocity is yet. In a sense, that is what you will end up solving for.]

Besides, it doesn't even take the mass of the disk into account.

Try to start part b) of the problem with conservation of angular momentum. (As an interim step, you'll need to find the disk's moment of inertia. )

 

[collinsmark]

By the way, that's one big frog!

 

[paisiello2]

You've assumed that the disk is rotating the same speed as the frog. Imagine if the disk was the planet Earth instead.

 

[Chase11]

Sorry I am still confused. I found the moment of inertia of the disk is I=1/2Mr^2 or (1/2)(4kg)(1.25m)^2=3.125kgm^2.

So if I use the conservation of angular momentum I have Iiωi=Ifωf. I don't understand what is going to change here between the initial and final values for the moment of inertia. Wouldn't it be the same for both?

 

[paisiello2]

Moment of inertia doesn't change and the angular momentum of the frog-disk system doesn't change either.

 

[Chase11]

Moment of inertia doesn't change and the angular momentum of the frog-disk system doesn't change either.

So how do I use Conservation of angular momentum to solve for this?

 

[paisiello2]

Before the frog jumped, what was the angular momentum of the system?

 

[Chase11]

Zero?

 

[paisiello2]

Correct, so what then is the angular momentum of the system after the frog jumped?

 

[Chase11]

It must also be zero.

 

[collinsmark]

Correct!

So you already know the angular momentum of the frog part of the frog-disk system (after the jump). What does that tell you about the disk part of the frog disk system (after the jump)?

 

[Chase11]

So is it 4.46 kgm^2/sec since the frog's was -4.46?

 

[collinsmark]

So is it 4.46 kgm^2/sec since the frog's was -4.46?

Right! Very nice.

So now, how does the disk's rotational period fit in?

 

[Chase11]

Okay so if I have ∠of disc=4.46kgm^2/sec 4.46=mvr

so 4.46=(4kg)ω(1.25m) I get ω=.892 m/s.

Then T=2pi/ω I get T=7.04sec. This is still the wrong answer.

 

[paisiello2]

What happened to your beautiful Iω that you spent all that time calculating?