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?

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

Answer 2

Well, it seems like your calculation is correct, but something seems to be missing. You forgot to factor in the moment of inertia of the rods themselves. You also forgot to account for the fact that the torque required to stop the balls will be in the opposite direction of the initial rotation.

So, let's correct that:

The inertia for each ball is given by: (1/2)*m*r^2 = (1/2)*(1.00kg)*(4.63m)^2 = 10.696995kg*m^2

The total inertia of the system is the sum of the inertia of the balls and the rods. Since the rods are massless, their inertia is negligible.

So, the total inertia is: 10.696995kg*m^2 + 10.696995kg*m^2 = 21.39399kg*m^2

Now, let's calculate the angular acceleration using the correct value for inertia:

Angular Acceleration = (Angular Velocity Final - Angular Velocity Initial)/time = (0 - (33rpm)(2πrad/1min))/4.07s

Angular Acceleration = -32.2122761rad/s^2

Finally, let's calculate the torque using the correct angular acceleration and inertia:

Torque = Angular Acceleration * Inertia = (-32.2122761rad/s^2)(21.39399kg*m^2) = -688.43252N*m

So, the correct torque required to bring the balls to a halt in 4.07s is -688.43252N*m (clockwise). Keep in mind that the negative sign indicates that the torque is in the opposite direction of the initial rotation.

Answer 3

Your calculations seem accurate. However, there's a small error in your answer at the end. The negative sign in front of the torque signifies that the torque is in the opposite direction of the initial rotation. So, the correct answer should be -12.1 Nm, not 12.1 Nm. This means that the torque needed to bring the balls to a halt is 12.1 N*m in the clockwise direction.

Answer 4

Your calculations are mostly correct, but there is a mistake in the calculation of the moment of inertia for the 2kg ball.

The correct calculation of the moment of inertia for the 2kg ball is:
Inertia = (2kg) * (1.5433m)^2 = 4.76924 kg*m^2

Now, let's recalculate the torque:
Angular Acceleration = (-(33.0 rpm) * 2π rad/rev) / (60s/min) / (4.07s) = -0.84908 rad/s^2

Torque = Angular Acceleration * Inertia
Torque = (-0.84908 rad/s^2) * (14.29085512 kg*m^2 + 4.76924 kg*m^2)
Torque = -15.51 N*m

So, the correct torque to bring the balls to a halt in 4.07s is -15.51 N*m.