--------My Work----------- First I converted 24 rpm to rads/s. (24*2pi)/60 = 2.5

Question

--------My Work-----------

First I converted 24 rpm to rads/s. (24*2pi)/60 = 2.513 rad/s

I tried to find the Angular Acceleration first using the equation omega final= omega initial +(alpha)(delta t). basically getting that the angular acceleration is

= (2.513 rad/s)/ 6s

=0.4188 rad/s^2

I also know that torque=I*(alpha)

So I tried to find the moment of inertia I using (m1 +m2)r^2

(1+2kg)0.4^2 =0.48 Kg*m^2 with r coming from d/2=(0.8/2)

So when I try to find the Torque by plugging everything in I get,

=(0.48kg*m^2)(0.4188 rad/s^2)

=0.201 N*m But Mastering Physics says I am wrong. Can someone do this problem and tell me what I am doing wrong?

A 1.0 kg s? --------My Work----------- First I converted 24 rpm to rads/s. (24*2pi)/60 = 2.513 rad/s I tried to find the Angular Acceleration first using the equation omega final= omega initial +(alpha)(delta t). basically getting that the angular acceleration is = (2.513 rad/s)/ 6s =0.4188 rad/s^2 I also know that torque=I*(alpha) So I tried to find the moment of inertia I using (m1 +m2)r^2 (1+2kg)0.4^2 =0.48 Kg*m^2 with r coming from d/2=(0.8/2) So when I try to find the Torque by plugging everything in I get, =(0.48kg*m^2)(0.4188 rad/s^2) =0.201 N*m But Mastering Physics says I am wrong. Can someone do this problem and tell me what I am doing wrong? rpm. What torque will bring the balls to a halt in 6.0s m-long rigid, massless rod. The rod is rotating cw about its center of mass at 24rpm kg ball are connected by a 0.80-m kg ball and a 2.0 kg

Explanation / Answer

First you need to find the center of mass. To do this use the equation:
MR = m1r1 + m2r2

Let's just say that baseball 1 is at a position r = 0, then baseball 2 would be at r = 1m
So, MR = 1kg * 0m + 2kg * 1m
MR = 2 kg*m
Then to find the position of the center of mass simply divide by the total mass (M, which is 1kg + 2kg = 3kg)
R = (2kg*m) / 3kg
R = 2/3 m = 'center of mass'

this makes sense too because we would expect the center of mass to be a little bit closer to the heavier ball.

Now, if it is rotating at 24rpm (rotations per minute) that means it is rotating at 24 * 2(pi) = 48pi radians per minute
which is equal to:
2.51 rads/s

That is the angular velocity, if we want it to stop in 6 seconds use this equation to find the acceleration:
acceleration = (Vf - Vi) / time
acceleration = (0rads/s - 2.51rads/s) /6s
acceleration = -0.418 rads/s^2
NOTE: this is angular acceleration, usually denoted with the character "alpha", and velocity is actually 'angular velocity' with the character "omega". I used V to make it easier, as I cannot type an "omega" sign.


Then to find the torque we use the equation:
torque = (moment of intertia) * (angular acceleration)
but we don't know what the moment of inertia is, so have to calculate it, this is done with the equation: (moment of inertia is denoted with the capital letter i)
I = m1*r1^2 + m2*r2^2 + m3*r3^2 + .......
taking the center of mass (2/3m measured from the 1kg mass) to be r = 0:
I = 1kg * (-2/3m)^2 + 2kg * (1/3m)^2
I = 2/3 kg*m^2

Then using the equation state above (torque = I * a)

torque = 2/3kg*m^2 * -0.418rads/s^2

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