Question

a 0.60-kg puck revolves at 2.4 m/s at the end of a 0.90-m string on a frictionless air table. if the string is pulled in until its length is 0.60 m, what is the new speed of the puck?

Answer 1

3.6 m/s

Step-by-step explanation:

The mass of the puck, m = 0.60kg

The speed of the puck, v1 - 2.4 m/s

The initial length of the string, r1 = 0.90 m

Formula Used:

The angular momentum of mass m moving in a circle or radius r with a speed v is given by,

L=mvr

And, when no external acts on a system, then the angular momentum of the system remains conserved.

Proof:

Since no external torque is exerted on the puck, hence the angular momentum of the puck string remains conserved.

Then, using the given values,

L1 = L2

mv1r1 = mv2r2

v2 = v1r1/r1

Using the given values in above,

v2 = (2.4)(0.90)/0.60

= 3.6 m/s

Conclusion:

Hence, when the string is shortened to 0.60m, the speed of the considered puck will be 3.6 m/s.