A yo-yo is made of two discs separated by a cylindrical axle. Thin string is wrapped tightly around the axle.
Initially, both the free end A\text{A}A of the string and the yo-yo are held stationary.
With A\text{A}A remaining stationary, the yo-yo is now released so that it falls vertically. As the yo-yo falls, the string unwinds from the axle so that the yo-yo spins about its centre of mass.
The linear velocity vvv of the centre of mass of the falling yo-yo is related to the angular velocity ω\omegaω by v=rωv = r\omegav=rω where rrr is the radius of the axle.
The yo-yo accelerates uniformly as it falls from rest. The string remains taut and has negligible thickness.
When the yo-yo has fallen a distance of 0.80 m0.80\text{ m}0.80 m, its linear velocity is VVV.
Calculate VVV by considering the energy transfers that occur during the fall.
The yo-yo falls further until all the string is unwound. The yo-yo then 'sleeps'. This means the yo-yo continues to rotate in a loose loop of string.
The string applies a constant frictional torque of 9.0×10−4 N m9.0 \times 10^{-4}\text{ N m}9.0×10−4 N m to the axle.
The angular velocity of the yo-yo at the start of the sleep is 120 rad s−1120\text{ rad s}^{-1}120 rad s−1.
Determine, in rad\text{rad}rad, the total angle turned through by the yo-yo during the first 8.0 s8.0\text{ s}8.0 s of sleeping.
