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morpeh [17]
4 years ago
11

A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.0

700 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
Physics
2 answers:
larisa86 [58]4 years ago
8 0
Final angular velocity = initial angular velocity plus the product of angular acceleration and time 

<span>w = wo + at </span>

<span>( 1/2 ) wo = wo + at </span>
<span>- ( 1/2 ) wo = at </span>
<span>- ( 1/2 ) ( 88 rad / s ) = a ( 4.40 s ) </span>
<span>a = -10 rad /s </span>

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

<span>Fr = I a </span>
<span>F ( .0700 m ) = ( .850 kg m^2 ) ( -10 rad / s ) </span>
<span>F = 120 N</span>
Tanya [424]4 years ago
4 0

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

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