A yoyo with a mass of m = 170 g is released from rest as shown in the figure. Th

Question

A yoyo with a mass of m = 170 g is released from rest as shown in the figure.

The inner radius of the yoyo is r = 2.32 cm, and the outer radius is R = 4.81 cm, and the moment of inertia about the axis perpendicular to the plane of the yoyo and passing through the center of mass is ICM = 1.68×10-4 kgm2. Determine the linear acceleration of the yoyo.

Determine the angular acceleration of the yoyo.

What is the weight of the yoyo?

What is the tension in the rope?

If a 1.54 m long section of the rope unwinds from the yoyo, then what will be the angular speed of the yoyo?

Explanation / Answer

A) Let the tension be T, acceleration be a, angular acceleration be a/r

Now force equation : ma = mg - T

Torque equation : i alpha = Tr or i a/r^2 = T

Adding the two equations

ma + i a/r^2 = mg

0.17a + 1.68e-4/0.0232^2 a = 0.17*9.8

0.482a = 0.17*9.8

a = 0.17*9.8/0.482

= 3.456 m/s^2

B) angular acceleration = a/r = 3.456/0.0232

= 149 rad/s^2

C) weight = mg = 0.17*9.8 = 1.666 N

D) T = ia/r^2

= 1.68e-4 *3.456/0.0232^2

= 1.08 N

E) v = sqrt(2as)

= sqrt (2*3.456*1.54)

= 3.26 m/s

Angular velocity = v/r = 3.26/0.0232

= 140.5 rad/s answer

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