Kdar1987 Posted November 26, 2007 Share Posted November 26, 2007 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? Link to comment Share on other sites More sharing options...
swansont Posted November 26, 2007 Share Posted November 26, 2007 You get 72 how? (i.e. show your work) What is exerting the torque? Link to comment Share on other sites More sharing options...
Kdar1987 Posted November 26, 2007 Author Share Posted November 26, 2007 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: Link to comment Share on other sites More sharing options...
swansont Posted November 26, 2007 Share Posted November 26, 2007 What's the angle? Link to comment Share on other sites More sharing options...
doG Posted November 26, 2007 Share Posted November 26, 2007 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? Link to comment Share on other sites More sharing options...
Kdar1987 Posted November 26, 2007 Author Share Posted November 26, 2007 You just restate the problem.. But how do I put 0.75m into the equation? I don't understand. Link to comment Share on other sites More sharing options...
doG Posted November 26, 2007 Share Posted November 26, 2007 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... Link to comment Share on other sites More sharing options...
Kdar1987 Posted November 26, 2007 Author Share Posted November 26, 2007 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 Link to comment Share on other sites More sharing options...
doG Posted November 26, 2007 Share Posted November 26, 2007 Congratulations. Link to comment Share on other sites More sharing options...
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