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

 

[rwisz]

Your process is correct, but your calculation for the moment of inertia (I) is incorrect. The moment of inertia for a rod rotating about its center of mass is given by I = (1/12)*m*L^2, where m is the mass of the rod and L is the length of the rod. In this case, the rod is 1.31 m long and has a mass of 3.15 kg, so the moment of inertia is (1/12)*(3.15)*(1.31)^2 = 0.343 kg*m^2. The moment of inertia for the 6.30 kg ball is negligible compared to the rod, so it can be ignored in this calculation.

To find the correct distance for the center of mass, you can use the equation for the center of mass of a system of two particles connected by a rigid rod:

r_cm = (m1*r1 + m2*r2)/(m1+m2)

In this case, m1 = 3.15 kg, m2 = 6.30 kg, r1 = 0 (since the rod's center of mass is at the origin), and r2 = 1.31/2 = 0.655 m (since the center of mass of the 6.30 kg ball is at the end of the rod). Plugging these values into the equation, we get r_cm = (3.15*0 + 6.30*0.655)/(3.15+6.30) = 0.393 m. This is the distance from the center of mass of the system to the axis of rotation.

Now, we can use the correct moment of inertia and the calculated distance to find the torque:

Torque = I*alpha = (0.343)*(2.53) = 0.867 N*m

This is the torque needed to bring the balls to a halt in 6.97 s.