Question

A 1.00kg ball and a 2.00kg ball are connected by a 4.63m long rigid, massless rod. The rod is rotating clockwise about its center of mass at 33.0rpm. What torque will bring the balls to a halt in 4.07s?

My Attempt:

Center of Mass = [(1kg)(0m) + (2kg)(4.63m)]/(1kg+2kg)
=9.26/3
=3.0866m

Intertia = sum of m*r^2
= ((1kg)(3.0866m)^2)+((2kg)(1.5433^2))
=14.29085512 kg*m^2

Angular Acceleration = [(angular velocity final) - (angular velocity initital)]/time

= [[-(33rpm)(2Pie)]/60sec]/(4.07sec)
= -0.849079095rad/s^2

Torque = Angular acc * Intertia

= (-0.849079095rad/s^2)(14.29085512kg*m^2)
= -12.1N*m
or
= 12.1 N*m ( counter clockwise)

Where am I going wrong?

Answer 1

n=33rpm=33/60 =0.55rev/sec
ω =2πn=2π•0.55=3.45 rad/s

The rod is rotating clockwise about ITS CENTER OF MASS.
The center of mass is at r=4.63/2 =2.315 m from each end of the rod.Then
M=Iε =(m1•r²+m2•r²)•ω/t = r²(m1+m2) •ω/t=2.315²•3•3.45/4.07 =13.63 N•m

If the system “two balls+massless rod” is rotating about the CENTER OF MASS OF THE SYSTEM, we have to find the CM.
The center of mass and the moment of inertia are
x(CM)= {m1•0 +m2•4.63)/ (m1+m2) = 2•4.63/3 =3.087 m,

I=1•3.087² +2(4.63-3.087)² = 9.53 +4.76=14.3 kg•m²
M=Iε =I •ω/t= 14.3•3.45/4.07 =12.12 N•m