Rotational Motion and Torque: Solving for Angular Acceleration

[Brucezhou]

The question is the first part of Question 63. It is required to make a proof. What I was trying to do is to build a equation with Torque= Moment of Inertia * angular acceleration. But when it was simplified to 3g cos(sita)= r* dw/dt, it seems not easy to solve it, since both of sita and dw/dt are not constant. What can I do next?

 

[ehild]

How did you get your equation? It is not correct.

You have to prove that the vertical component of the acceleration of the end of the rod is greater than the acceleration of the ball. ehild

 

[Brucezhou]

Let Wr cos(sita)=1/3 mr^2* dw/dt.
Simplify it then get:

3g*cos(sita)=r*dw/dt

 

[Brucezhou]

How did you get your equation? It is not correct.

You have to prove that the vertical component of the acceleration of the end of the rod is greater than the acceleration of the ball. ehild

Could you suggest a way to solving the problem? I don't have a standard answer key and I don't know where I got wrong. I have published my method above

 

[ehild]

Let Wr cos(sita)=1/3 mr^2* dw/dt.
Simplify it then get:

3g*cos(sita)=r*dw/dt

What is W?

ehild

 

[ehild]

Could you suggest a way to solving the problem? I don't have a standard answer key and I don't know where I got wrong. I have published my method above

The answer is written in the problem text. You do not need to solve the differential equation. As both the ball and the rod start from rest, you only need to show R what condition is the acceleration of the released ball smaller then the vertical acceleration of the end of the rod.
First you need the correct expression for the torque.

ehild

 

[Brucezhou]

What is W?
ehild

weight

 

[Brucezhou]

The answer is written in the problem text. You do not need to solve the differential equation. As both the ball and the rod start from rest, you only need to show R what condition is the acceleration of the released ball smaller then the vertical acceleration of the end of the rod.
First you need the correct expression for the torque.

ehild

Thank you a lot. I have solved. First my Torque equation is wrong. It should be 1/2 WL cos(sita) since the weight is exerted in the middle. Then it should be
2/3*a(tangent)=g*cos(sita)
2/3*cos(sita)*a(tangent)=g*cos^2(sita)
cos(sita)*a(tangent)=a(vertical to ground)
which should be at least equal g
so
2/3=cos^2(sita)
sita=cos-1(squr(2/3))

 

[ehild]

It is correct now.
You can improve the formatting to hit the button "go advanced". On the right, you find Greek letters like θ (theta) or mathematical symbols like √ (square root) . Above you see the symbols x² and x₂: you can use x² to write power (3²=9) or inverse function θ=cos^-1(2/3). x₂ is used when indexing quantities. For example, when you write ∑x_i=0. Explore!

ehild