Question

A pair of constant forces of magnitude F = 18.3 N is applied to opposite sides of the axle of a top, as shown in the figure. The diameter of the axle is d = 8.85 mm. The angle θ, 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?

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. If this helps then kindly rate 5-stars

A pair of constant forces of magnitude F = 18.3 N is applied to opposite sides of the axle of a top, as shown in the figure. The diameter of the axle is d = 7.95 mm. The angle theta, which has a value of 32.1 degree, describes the steepness of the top's sloping sides. The moment of inertia of the top about its spin axis is l = 0.675 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?

solution:

T = I alpha = FL

In this case, two 12.7 N forces are acting on an axle

(2)(18.3)(3.975 * 10-3) = (0.675) alpha

alpha = 0.216 rad/s2

a = alpha r

Using the geometry of the picture, the radius is the opposite side of the triangle formed by the given angle

tan theta = r / h

tan 32.1 = r / 5.13

r = 3.22 cm

a = alpha r = (0.216)(0.0322)

a = 6.96 * 10-3 m/s2 = 0.696 cm/s2