Angular Acceleration Problem/ Solid Cylinder

[melissa_y]

1) A solid cylinder of mass M = 1.14 kg and radius R = 6.3 cm pivots on a thin, fixed, frictionless bearing (see Figure). A string wrapped around the cylinder pulls downward with a force F = 4.606 N. What is the magnitude of the angular acceleration of the cylinder? Use units of "rad/s\^{}2".

2)Consider that instead of the force F, a block with mass m = 0.47 kg (with force F = 4.606 N) is attached to the cylinder with a massless string (see Figure). What is now the magnitude of the angular acceleration of the cylinder? Use units of "rad/s\^{}2".

3) How far does mass m travel downward between t = 0.51 s and t = 0.71 s (Assuming motion begins at time t = 0.0 s)?

 

[member 553083]

Use units of "m".Answer:1) The magnitude of the angular acceleration of the cylinder is 0.50 rad/s^2.2) The magnitude of the angular acceleration of the cylinder is 0.76 rad/s^2.3) The mass m travels a distance of 0.20 m downward between t = 0.51 s and t = 0.71 s.

 

[thepatientmental]

1) To solve this problem, we can use the formula for angular acceleration: α = τ/I, where α is the angular acceleration, τ is the torque, and I is the moment of inertia. In this case, the torque is caused by the force F pulling downward on the string, and the moment of inertia for a solid cylinder is given by I = 1/2MR^2. Plugging in the values, we get:

α = (4.606 N * 6.3 cm)/ (1/2 * 1.14 kg * (6.3 cm)^2)
= 3.64 rad/s^2

Therefore, the magnitude of the angular acceleration of the cylinder is 3.64 rad/s^2.

2) In this case, the block attached to the cylinder adds to the total mass and changes the moment of inertia. The new moment of inertia is given by I = 1/2(M + m)R^2. Plugging in the values, we get:

α = (4.606 N * 6.3 cm)/ (1/2 * (1.14 kg + 0.47 kg) * (6.3 cm)^2)
= 2.99 rad/s^2

Therefore, the magnitude of the angular acceleration of the cylinder is 2.99 rad/s^2.

3) To find the distance traveled by the block, we can use the formula for displacement: s = ut + 1/2at^2, where s is the displacement, u is the initial velocity (which is zero in this case), a is the acceleration (which we can find using the formula α = F/mR), and t is the time interval.

Plugging in the values, we get:

s = 0 + 1/2 * (4.606 N/0.47 kg * 6.3 cm) * (0.71 s)^2 - 1/2 * (4.606 N/0.47 kg * 6.3 cm) * (0.51 s)^2
= 0.028 m

Therefore, the block travels a distance of 0.028 m between t = 0.51 s and t = 0.71 s.