- #1
cdbowman42
- 14
- 0
1. A 2.5 kg block and a 1.5 kg block are attached to opposite ends of a light rope. The rope hangs over a solid frictionless pulley that is 30 cm in diameter and has a mass of .75 kg. When the blocks are released, what is the acceleration of the lighter block.
Ok, the book says 2.2m/s^2, but i cannot come to that, and the book has been wrong in the past. I'm pretty sure the acceleration on the block will be the same as the tangential acceleration of the rope over the pulley, which is angular acceleration*r
2. a=angular acceleration/r
angular accereration=Net Torque/Moment of Inertia(I)
Net Torque=mass of b1*gr-mass of b2*gr
I=1/2(mass of pulley)(r)^2
3. Ok here we go.
I=1/2(.75kg)(.15)^2
=0084375kg*m^2
Net Torque=2.5kg(9.8m/s^2)(.15m)-1.5kg(9.8m/s^2)(.15m)
=3.675N*m-2.205N*m
=1.47N*m
angular acceleration= 1.47/.0084375
=174.22rad/s^2
a=174.22rad/s^2(.15m)
=26.133m/s^2
Who's right and who's wrong, or are both of us wrong?
Ok, the book says 2.2m/s^2, but i cannot come to that, and the book has been wrong in the past. I'm pretty sure the acceleration on the block will be the same as the tangential acceleration of the rope over the pulley, which is angular acceleration*r
2. a=angular acceleration/r
angular accereration=Net Torque/Moment of Inertia(I)
Net Torque=mass of b1*gr-mass of b2*gr
I=1/2(mass of pulley)(r)^2
3. Ok here we go.
I=1/2(.75kg)(.15)^2
=0084375kg*m^2
Net Torque=2.5kg(9.8m/s^2)(.15m)-1.5kg(9.8m/s^2)(.15m)
=3.675N*m-2.205N*m
=1.47N*m
angular acceleration= 1.47/.0084375
=174.22rad/s^2
a=174.22rad/s^2(.15m)
=26.133m/s^2
Who's right and who's wrong, or are both of us wrong?