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At the instant r = 1.5 m , the 7-kg disk is given a speed of v = 5 m/s , perpendicular to the elastic cord. The disk slides on the smooth horizontal plane. Neglect its size. The cord has an unstretched length of 0.5 m .
(Figure 1)

Part A

Determine the speed of the disk at the instant r = 1.2 m .

Express your answer to three significant figures and include the appropriate units.

Part B

Determine the rate of shortening of the elastic cord at the instant r = 1.2 m .

Express your answer to three significant figures and include the appropriate units.

Part a =6.29 m/s im only lookin for part B please make sure its correct cus i got one shot only on mastering.

Explanation / Answer

Part B) As, torque = Force * radius

                         => Torque =   moment of inertia * angular aceleration

Here,   Wf = Wi + alpha * t

=>     rate of shortening of the elastic cord = 6.29 * (1.2/1.5)

                                                                  =   5.032   m/sec

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