Question

Block A (0.40 kg) and block B (0.30 kg) are on a frictionless table. Spring 1 connects block A to a frictionless peg at 0 and spring 2 connects block A and block B. When the blocks are in uniform circular motion about 0, the springs have lengths of 0.60 m and 0.40 m, as shown. The springs are ideal and massless, and the linear speed of block B is 2.0 m/s. If the spring constant of spring 1 is equal to 30 N/m, the unstretched length of spring 1 is closest to?

Answer 1

Since Block A and block B are connected, they have the same angular velocity.
Angular velocity = linear velocity � radius
For block B, the radius is 1.0 meter.
Angular velocity = 2 � 1 = 2 rad/s

For block A, the radius is 0.6 meter.
2 = linear velocity � 0.6, Linear velocity = 1.2 m/s
This is the speed of block A.

Centripetal force = m * v^2/r
Since the springs have lengths of 0.60 m and 0.40 m, the distance from the peg to the block B is 1.0 meter. Since Block B is moving around a circle at a speed of 2 m/s, let�s determine the centripetal force.
Fc = 0.30 * 2^2/ 1 = 1.2 N
This is the force which spring 2 is exerting on block B.

Block A has two forces. Spring 2 is pulling is exerting a 1.2 N force, which is pulling it away from the peg. Spring 1 is pulling it toward the peg.
Net force toward the peg = Force of Spring 1 � 1.2

Block A is moving around a circle with a radius of 0.6 meter.
For block A, Fc = 0.4 * 1.2^2/0.6 = 0.96 N
This is the net force on block A.

This force caused Spring 1 to stretch from its original length to a length of 0.6 meter

F = k * ? d
0.96 = 30 * ? d
? d = 0.96 � 30
Original length = 0.6 � (0.96 � 30) = 0.568 meter