Question

A yoyo is used to perform “the sleeper” trick. The yoyo is thrown downward and is made to spin at

the end of its string.The inertia of the yoyo is I = 2.4 × 10−5kg · m2
(a) The angular momentum magnitude for the yoyo is 6 × 10−4 kg · m2/s. What is the angular velocity of
the yoyo? Remember, this is a vector quantity.
(b) There is some friction between the yoyo and its string. If this applies a torque (magnitude) of
3 × 10−5 N · m, how long can the trick be performed before the yoyo stops spinning

Answer 1

The question seeks to determine the time duration for the 'sleeper' yoyo trick with a given torque, which involves understanding torque, angular momentum, and friction. However, essential details such as the opposing frictional torque or the yoyo's moment of inertia are missing to provide a precise answer.

Step-by-step explanation:

The student is asking about the duration a yoyo can perform the sleeper trick given a certain amount of torque. The calculation involves physics concepts such as torque, angular momentum, and friction.

However, the actual question seems to be missing some necessary details such as the frictional torque opposing the spinning or the moment of inertia of the yoyo. Generally, the time duration for which the yoyo can spin depends on its initial angular momentum and the torque applied by friction which causes it to decelerate. Without the opposing torque or angular momentum, we cannot accurately calculate the time it will take for the yoyo to stop spinning.