block 1 has mass m1 = 480 g, block 2 has mass m2 = 540 g, and the pulley is on a frictionless horizontal axle and has radius R = 5.2 cm. When released from rest, block 2 falls 76 cm in 5.2 s (without the cord slipping on the pulley). What is the pulley's rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s2.
h=at²/2
a=2h/t²=2•0.76/5.2²=0.056 m/s²
Two equations for the loads
-m₁a=m₁g-T₁ => T₁=m₁(g+a) ….(1)
m₂a=m₂g-T₂,=> T₂=m₂ (g+a) ….(2)
and the 3rd equation (on the base of Newton’s 2 law for rotation Iε=M)
I•a/R=( T₂-T₁)R …. (3)
Substitute (1) and (2) in (3)
and solve for I
To find the pulley's rotational inertia, we need to use the information given and apply some principles of physics.
1. Start by determining the acceleration of block 2 using the formula for acceleration:
a = (vf - vi) / t
Here, vf (final velocity) can be taken as 0 since block 2 falls vertically, and vi (initial velocity) is unknown. Thus, the formula becomes:
a = (0 - vi) / t
Rearranging the formula:
vi = - a * t
Substituting the given values:
vi = - (9.81 m/s^2) * (5.2 s)
2. Now, we can determine the net force acting on block 2. As block 2 falls, the gravitational force pulls it downward, and the tension in the cord pulls it upward. The net force can be calculated using Newton's second law:
F_net = m2 * a
Substituting the given values:
F_net = (0.540 kg) * (-9.81 m/s^2)
3. The net force can also be calculated in terms of the tension provided by the cord:
F_net = 2 * T
(Since the force provided by the tension in the cord is equal and opposite to the gravitational force on block 2)
4. Equating the two expressions for net force:
2 * T = m2 * a
Rearranging the equation:
T = (m2 * a) / 2
Substituting the values:
T = (0.540 kg) * (-9.81 m/s^2) / 2
5. Now, let's calculate the torque exerted by the tension force on the pulley. The torque can be defined as the product of force and the distance from the axis of rotation (radius):
torque = T * R
Substituting the values:
torque = ((0.540 kg) * (-9.81 m/s^2) / 2) * (0.052 m)
6. Finally, we have found the torque exerted on the pulley. The rotational inertia of the pulley (I) can be calculated using the formula:
torque = I * angular acceleration
However, since the pulley is at rest initially, the angular acceleration is zero (as the cord does not slip on the pulley).
Therefore, the equation becomes:
torque = I * 0
Which simplifies to:
torque = 0
Hence, the torque exerted on the pulley is zero.
Therefore, the rotational inertia of the pulley is zero since there is no torque exerted on it.