A 2.0 kg stone is tied to a 0.50 m string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is:

 

 

 

 

so. t=rmasin(angle) or t=r^2*m*w^2*sin(angle)

 

I get 72.. Got the answer should be 108.

 

But.. I don't get how to get the torque about the origin? how do I use 0.75 m on Z axis?

You get 72 how? (i.e. show your work) What is exerting the torque?

You get 72 how? (i.e. show your work) What is exerting the torque?

 

 

t=r^2*m*w^2*sin(angle)

 

t=0.50^2*2.0*12^2*sin(angle)

 

A 2.0 kg stone is tied to a 0.50 m string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is:

What's the angle?

Picture a pole, like a flagpole, with a string attached .75m from the ground. There's a 2kg stone attached to the string spinning around the pole at 12 rad/s which exerts a centrifugal force on the pole .75m above its base. What is the resulting torque that pole exerts on it's base?

You just restate the problem..

 

But how do I put 0.75m into the equation? I don't understand.

You just restate the problem..

 

But how do I put 0.75m into the equation? I don't understand.

 

The .75m is the length of a lever where the force applied is that resulting from the centrifugal force of the stone spinning around that lever. Imagine that you are holding a stick with a stone attached to a string .75m from your hand. If you spin the stone about the stick it will exert a force on your hand.

 

BTW, the formula you are using is not the one to calculate the force that results from the orbiting stone...

oh.. Ok. I think I getting it now.

 

So.. I need to find rotational force of the ball? and then multiply by 0.75m?

 

F= m*a = m*w^2*r =2*12^2*0.5 = 144

 

t = F*d = 144*075 = 108

 

Right?

 

Thnks by the way

Congratulations.