A solid, frictionless cylindrical pulley has a radius R = 0.600 m. A block is at

Question

A solid, frictionless cylindrical pulley has a radius R = 0.600 m. A block is attached to a cord that is wrapped around the outside of the pulley. The block is falling with a downward acceleration of 5.00 m/s2, and the tension in the cord is 0.850 N. The free body diagrams for the pulley and the block are shown below. Note that the magnitudes of the two tensions 7j and T2 are equal. (Just call the tension T.) Also, "n" in the free body diagram of the pulley represents the normal force provided by the axle of the pulley, equal in magnitude to the weight of the pulley. Note that the normal force and the weight force do not produce torques on the pulley. (Why is this?) Determine the angular acceleration of the pulley.rad/s2 Determine the moment of inertia of the pulley.

Explanation / Answer

The normal force and the weight force did not produce a torque because they are not acting on the pulley in the direction of a tangent to the pulley’s surface.

a) angular acceleration = linear acceleration a / radius of gyration R

= 5 / 0.6

= 8.333 rad/s2

d)

to find the mass of the block m

ma = mg-T

mx5 = mx9.8 – 0.85

0.85 = (9.8 – 5)m

m = 0.177 kg

b) moment of inertia of the pulley

torque = ma x R = I

I = maR/ = maR2/a = mR2

I = 0.177 x 0.62

I = 0.06372 kgm2

= 0.531 N

C)

mg – T = ma

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.