Initially, a small 2.0-kg rock is whirling at the end of a very thin string in a circular path of radius 0.75 m on a horizontal frictionless surface, as shown in the figure. The initial tangential speed of the rock was 5.0 m/s. The string has been slowly winding around a vertical rod, and a few seconds later the length of the string has shortened to 0.25 m. What is the instantaneous speed of the mass at the moment the string reaches a length of 0.25 m?
A) 3.9 m/s
B) 9.3 m/s
C) 15 m/s
D) 225 m/s
E) 75 m/s

Question

Initially, a small 2.0-kg rock is whirling at the end of a very thin string in a circular path of radius 0.75 m on a horizontal frictionless surface, as shown in the figure. The initial tangential speed of the rock was 5.0 m/s. The string has been slowly winding around a vertical rod, and a few seconds later the length of the string has shortened to 0.25 m. What is the instantaneous speed of the mass at the moment the string reaches a length of 0.25 m? 
A) 3.9 m/s 
B) 9.3 m/s 
C) 15 m/s 
D) 225 m/s 
E) 75 m/s

Quizplus · Accepted Answer

The angular momentum of the system remains constant since there is no external torque acting on it. 
$L = mvr$
Where L = angular momentum, m = mass, v = tangential velocity, and r = radius of circular path.

Initial angular momentum = $2.0\ kg \cdot 5.0\ m/s \cdot 0.75\ m = 7.5\ kg\cdot m^2/s$

Final angular momentum = $2.0\ kg \cdot v_f \cdot 0.25\ m$

Setting the two equal to each other and solving for $v_f$, we get:

$v_f = \frac{7.5\ kg\cdot m^2/s}{2.0\ kg \cdot 0.25\ m} = 15\ m/s$

Therefore, the instantaneous speed of the rock at the moment the string reaches a length of 0.25 m is 15 m/s.

Initially, a Small 2