A cylinder is rotating about an axis that passes through the center of each circ

Question

A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.0850 m, an angular speed of 88.0 rad/s, and a moment of inertia of 0.850 kg · m2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of 4.20 s.

(a) Find the magnitude of the angular deceleration of the cylinder.


(b) Find the magnitude of the force of friction applied by the brake shoe.

Explanation / Answer

a)

final angular velocity = initial angular velocity plus the product of angular acceleration and time

w = wo + at

( 1/2 ) wo = wo + at

- ( 1/2 ) wo = at

- ( 1/2 ) ( 88 rad / s ) = a ( 4.20 s )

a = -10.48 rad /s

Newton's Second Law, rotational form: Torque (force perpendicular to radius) is equal to the product of moment of inertia and angluar acceleration

Fr = I a

F ( 0.0850 m ) = (0 .850 kg m^2 ) ( -10.48 rad / s )

F = 104.8 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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