Question 4 of 12 Map sapling earning A pair of constant forces of magnitude F 12

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Question 4 of 12 Map sapling earning A pair of constant forces of magnitude F 12.7 N is applied to opposite sides of the axle of a top, as shown in the figure. The diameter of the axle is d 8.13 mm. The angle 6, which has a value of 33.79, describes the steepness of the top's sloping sides. The moment of inertia of the top about its spin axis is 0.699 kg. m2 What is the tangential acceleration at of the point labeled P, which is at a height of h 5.13 cm above the floor? click to edit Number cm. s frictionless floor

Explanation / Answer

let r = d/2 = 8.13/2 = 0.004065 m

then torque = F * r = 2 * 12.7 N * 0.004065 m = 0.103251 N.m

but also torque = I*a,

so 0.103251 N.m = 0.699 kg.m^2 *

a = 0.1477 rad/s^2

applies to all points on the top.

tan(theta) = R / h

tan(33.7) = R / 0.0513m

R = 0.0342 m

so at point P, tangential acceleration = a*R = 0.1477 rad/s^2 * 0.0342m = 0.00505134m/s^2 = 5.05134 mm/s^2

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