A pair of constant forces of magnitude F = 12.7 N is applied to opposite sides o

Question

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 = 9.39 mm. The angle , which has a value of 33.7°, describes the steepness of the top's sloping sides. The moment of inertia of the top about its spin axis is I = 0.627 kg·m2. What is the tangential acceleration at of the point labeled P, which is at a height of h = 4.75 cm above the floor?

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 = 9.39 mm. The angle theta, which has a value of 33.7 degree, describes the steepness of the top's sloping sides. The moment of inertia of the top about its spin axis is I = 0.627 kg.m^2. What is the tangential acceleration at of the point labeled P, which is at a height of h = 4.75 cm above the floor?

Explanation / Answer

The formual for torque is Force times moment arm. Torque also equals I

So FL = I    L, the moment arm is the radius of the axle

(2)(12.7)(4.695 X 10-3) = (.627)

= .190 rad/s2

The tangential acceleration is found by a = r

We need to find the radius at the point in question. The radius is tohe opposite side of a triangle formed by the angle given and the height to that point.

tan = O/A

r = (.0475)(tan 33.7) = .0317 m

a = (.0317)(.190) = 6.02 X 10-2 m/s2

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