A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.0700 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 4.40 s.(a) Find the magnitude of the angular deceleration of the cylinder. rad/s 2 (b) Find the magnitude of the force of friction applied by the brake shoe. N
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 ( 4.40 s )
a = -10 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 ( .0700 m ) = ( .850 kg m^2 ) ( -10 rad / s )
F = 120 N
Answer:
-10 rad/s^2
121.43 N
Explanation:
Using rotational kinematics equation:
\(\alpha = \frac{w-w_{0} }{t} \\\\\alpha = \frac{44-88 }{4.4}\\\\\alpha = - 10rad/s^2\)
Using Newton's Second Law:
\(F*r = I*\alpha = sum of moments\\\\F = \frac{I*\alpha }{r} \\\\F = \frac{0.85*10}{0.07} \\\\F = 121.43 N\)