http://session.masteringphysics.com/prob…
The pulley in the figure has radius 0.160 m and a moment of inertia 0.480 kg*m^2 . The rope does not slip on the pulley rim.
How do I use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor?
The pulley in the figure has radius 0.160 m and a moment of inertia 0.480 kg*m^2 . The rope does not slip on the pulley rim.
How do I use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor?
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The pulley in the figure has radius 0.160 m and a moment of inertia 0.480 kg*m^2 . The rope does not slip on the pulley rim.
How do I use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor?
The rope is exerting a force on the outer edge of the pulley. The force is the difference of the weights of the blocks.
Net force on the pulley = 4 * 9.8 – 2 * 9.8 = 19.6 N
This force is doing 2 things at the same time.
1. The force is causing the 4 kg block to accelerate downward and the 2 kg block to accelerate upward.
2. The force is causing the pulley to accelerate as it rotates.
To determine the effect of the net force on the blocks and pulley at the same time, we have to determine the torque and the total moment of inertia.
Torque = Force * radius = 19.6 * 0.160 = 3.136 N * m
Moment of inertia of each block = mass * (perpendicular distance from the center of the pulley)^2
The horizontal distance between the block and the center of the pulley is the radius of the pulley.
Moment of inertia of 4 kg block = 4 * 0.16^2
Moment of inertia of 2 kg block = 2 * 0.16^2
Total moment of inertia = 0.480 + 6 * 0.16^2 = 0.6336
Torque = I * α
3.136 = 0.6336 * α
α = 4.949 radians per second^2
This is the angular acceleration of the pulley.
Since the rope does not slip, the linear acceleration of the rope = α * radius
Linear acceleration = 4.949 * 0.16 = 0.79184 m/s^2
This is the acceleration of rope and 4 kg block as the block moves 5 meters down!
Final velocity^2 = Initial velocity^2 + (2 * a * d)
Initial velocity = 0
Final velocity = (2 * 0.79184 * 5)^0.5 = 2.814 m/s
I have a program called Interactive Physics. I drew a simulation similar to this problem. The angular acceleration of the pulley = 4.924 rad/s^2
So, I believe that my answer is correct.
How do I use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor?
The rope is exerting a force on the outer edge of the pulley. The force is the difference of the weights of the blocks.
Net force on the pulley = 4 * 9.8 – 2 * 9.8 = 19.6 N
This force is doing 2 things at the same time.
1. The force is causing the 4 kg block to accelerate downward and the 2 kg block to accelerate upward.
2. The force is causing the pulley to accelerate as it rotates.
To determine the effect of the net force on the blocks and pulley at the same time, we have to determine the torque and the total moment of inertia.
Torque = Force * radius = 19.6 * 0.160 = 3.136 N * m
Moment of inertia of each block = mass * (perpendicular distance from the center of the pulley)^2
The horizontal distance between the block and the center of the pulley is the radius of the pulley.
Moment of inertia of 4 kg block = 4 * 0.16^2
Moment of inertia of 2 kg block = 2 * 0.16^2
Total moment of inertia = 0.480 + 6 * 0.16^2 = 0.6336
Torque = I * α
3.136 = 0.6336 * α
α = 4.949 radians per second^2
This is the angular acceleration of the pulley.
Since the rope does not slip, the linear acceleration of the rope = α * radius
Linear acceleration = 4.949 * 0.16 = 0.79184 m/s^2
This is the acceleration of rope and 4 kg block as the block moves 5 meters down!
Final velocity^2 = Initial velocity^2 + (2 * a * d)
Initial velocity = 0
Final velocity = (2 * 0.79184 * 5)^0.5 = 2.814 m/s
I have a program called Interactive Physics. I drew a simulation similar to this problem. The angular acceleration of the pulley = 4.924 rad/s^2
So, I believe that my answer is correct.