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.
To find the pulley's rotational inertia, we can use the information given about the blocks and the motion of block 2.
First, let's find the acceleration of block 2. We can use the kinematic equation: s = ut + (1/2)at^2.
Given:
Initial velocity, u = 0 (released from rest)
Distance, s = 76 cm = 0.76 m
Time, t = 5.2 s
Using the equation s = ut + (1/2)at^2, we can solve for acceleration, a:
0.76 = 0 + (1/2) * a * (5.2)^2
0.76 = (1/2) * a * 27.04
a = (2 * 0.76) / 27.04
a = 0.0562 m/s^2
Now, let's find the tension in the cord connecting block 2 to the pulley. We can use Newton's second law: ΣF = ma.
The net force acting on block 2 is the difference between its weight (mg) and the tension in the cord (T). So,
m2g - T = m2a
Substituting the given values:
0.540 kg * 9.81 m/s^2 - T = 0.540 kg * 0.0562 m/s^2
Simplifying further:
5.3046 N - T = 0.03 N
T = 5.2746 N
Next, we need to find the torque on the pulley. The torque exerted by the tension in the cord is given by T * R, where R is the radius of the pulley.
T = 5.2746 N
R = 5.2 cm = 0.052 m
Torque = T * R = 5.2746 N * 0.052 m = 0.2748 N*m
Finally, to find the pulley's rotational inertia (I), we can use the equation:
Torque = I * angular acceleration
Angular acceleration, α, can be found from the linear acceleration, a, using the equation:
α = a / R
Substituting the values:
α = 0.0562 m/s^2 / 0.052 m = 1.0808 rad/s^2
Now, rearranging the equation for torque:
I = Torque / α = 0.2748 N*m / 1.0808 rad/s^2
Calculating the result:
I = 0.254 kg*m^2
Therefore, the pulley's rotational inertia is 0.254 kg*m^2.