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

[Leeoku]

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?

 

[Filip Larsen]

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, x_cm, where m₁(x_cm-x₁) + m₂(x_cm-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 x_cm.

 

[James50]

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.

 

[Leeoku]

got it! thanks for the explanation on CM