Question

A yo-yo is made of two uniform disks, each of mass M and radius R, which are glued to a smaller central axle of negligible mass and radius 0.5R . A string is wrapped tightly around the axle. The yo-yo is then released from rest and allowed to drop downwards, as the string unwinds without slipping from the central axle.

A) Find the moment of inertia, I, of the yo-yo with respect to an axis through the common centers of the disks, in terms of the mass and radius.
B) Find the linear speed V of the yo-yo, after it has descended a distance H.
C) Calculate the magnitude of the linear velocity V, in meters per second, of the yo-yo after it has fallen a distance 0.33 m.

Answer 1

The moment of inertia, I, of the yo-yo with respect to an axis through the common centers of the disks is (13/8)MR^2. The linear speed, V, of the yo-yo after descending a distance H is √(16/13)(2gH). The magnitude of the linear velocity, V, of the yo-yo after falling a distance 0.33 m is 4.845 m/s.

Step-by-step explanation:

To find the moment of inertia, I, of the yo-yo with respect to an axis through the common centers of the disks, we can use the parallel-axis theorem. The moment of inertia of the disks about their centers is 1/2MR^2 each. The moment of inertia of the central axle is MR^2/16. Therefore, the total moment of inertia is I = 2(1/2MR^2) + MR^2/16 = 13/8MR^2.

To find the linear speed V of the yo-yo after it has descended a distance H, we can use the conservation of energy. The initial potential energy at the top of the descent is MgH. The final kinetic energy is (1/2)Iω^2, where ω is the angular velocity given by V/R. Equating the initial potential energy to the final kinetic energy, we have MgH = (1/2)(13/8MR^2)(V/R)^2. Solving for V, we get V = √(16/13)(2gH).

To calculate the magnitude of the linear velocity V of the yo-yo after it has fallen a distance 0.33 m, we can use the equation V = √(16/13)(2gH), where H is the distance fallen. Plugging in the values, we get V = √(16/13)(2 * 9.8 m/s^2 * 0.33 m) = 4.845 m/s.