How Much Torque Is Needed to Stop Rotating Balls in 6.97 Seconds?

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To determine the torque needed to stop two rotating balls connected by a rod, the calculation involves understanding angular speed and acceleration. The initial angular speed is given as 24.2 rpm, which converts to an angular acceleration of 2.53 rad/s². The correct torque is calculated using the moment of inertia and angular acceleration, resulting in a value of 10.3 N*m. The center of mass must be accurately determined, as it affects the torque calculation; the position along the rod can be found using the masses of the balls. Proper notation is crucial, distinguishing between angular speed and angular acceleration for clarity in calculations.
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Homework Statement


A 3.15 kg ball and a 6.30 kg ball are connected by a 1.31 m long rigid, massless rod. The rod is rotating clockwise about its center of mass at 24.2 rpm. What torque will bring the balls to a halt in 6.97 s?
Answer: 1.31e+00 N*m


Homework Equations


torque = I*alpha


The Attempt at a Solution


Alpha = 24.2*2pi/60
= 2.53
Torque = I*alpha
= 2.53(3.15*(1.31/2)^2+6.3(1.31/2)^2)
= 10.3

Ok so i think my process is right but my Centre of mass is probably wrong. is this because of the different weighting of both sides? Now i know Centre of mass = mr^2, but how do know how far each contributes?
 
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You are mixing things up a bit.

In your first equation you use alpha to signify angular speed, but in the second equation it signifies time rate change of angular speed, that is, angular acceleration. Following the usual notation you should probably use omega for angular speed in the first equation and then alpha for angular acceleration. You also need one more equation that makes the kinematic relationship between omega and alpha as a function of time.

You are correct that your center of mass is wrong. From definition of the CM you need to find the position along the rod, xcm, where m1(xcm-x1) + m2(xcm-x2) = 0 with x values being distances with sign along the rod from some reference point. For instance, choosing the m1 end of the rod as reference you have x1 = 0 and x2 = length of rod and you can easily solve for xcm.
 
The centre of mass is where the rod would balance perfectly on the point of a pin. To achieve this the torques on each side must be equal. Torque = Force x Distance. So the distance to one of the balls (from the centre of mass), multipled by that weight, must equal the distance to the other ball (from the centre of mass), multiplied by that weight. Then remember those two distances must add up to the full length of the rod. Hope that helps.
 
got it! thanks for the explanation on CM
 
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