Question

A 2.0 kg ball and a 3.5 kg ball are connected by a 3.0-mm-long rigid, massless rod. The rod and balls are rotating clockwise about its center of gravity at 18 rpm.

What magnitude torque will bring the balls to a halt in 6.0 s?

Answer 1

Here we have given that,

m1= 2.0 kg ball and

m2= 3.5 kg ball

Length of rod = 3mm

Speed = 18 rpm.

So that for the magnitude torque will bring the balls to a halt in 6.0 s we have,

For the center of mass we have,

MR = m1r1 + m2r2

Let's just say that ball 1 is at a position r = 0, then baseball 2 would be at r = 3 m

So, MR = 1kg * 0m + 3.5kg * 3m

MR = 10.5

Here ,

(M, =  2kg + 3.5kg = 5.5kg)

R = (10.5kg*m) / 5.5 kg

R = 1.91 m = ' which is the center of mass'

Now, if it is rotating at 18rpm that means it is rotating at 18* 2(pi) = 36pi radians per minute = 1.88495559215  rads/s which is the the angular velocity

Now if we want it to stop in 6 seconds then using

acceleration = (Vf - Vi) / time

acceleration = (0 -  1.88) / 6

acceleration = -0.314159 rads/s^2

Now we know that,

torque = (moment of intertia) ×(angular acceleration)

And for moment of inertia we will have

I = m1*r1^2 + m2*r2^2

taking the center of mass (1.91m measured from the 2kg mass) to be r = 0: so that

I = 2kg * (-1.91m)^2 + 3.5kg * (1.09zm)^2

I = 11.4545 kg*m^2

Then using the equation state above we will have,

Magnitude of the torque will be,

torque = 11.4544 × 0.314159

torque = 8.6025 Nm

Hence the torque will be 8.6025 Nm