Question

the 4.00 kg block is attached to a vertical rod by means of two strings. when the system rotates about the axis of the rod, the strings are extended as shown in the (figure 1) and the tension in the upper string is 82.0 n . find the number of revolutions per minute at which the lower cord just goes slack.

Answer 1

To find the number of revolutions per minute at which the lower cord goes slack, we need to analyze the forces acting on the block. When the block is rotating, the tension in the upper string creates a centripetal force, pulling the block towards the rod. The tension in the lower string is equal to the gravitational force acting on the block, which is equal to its weight. The lower string goes slack when the tension in the lower string is equal to zero, which happens when the centripetal force equals the weight of the block. To find the number of revolutions per minute, we need to use the following equation:

T = m * r * ω^2

where T is the tension in the lower string (equal to the weight of the block), m is the mass of the block (4.00 kg), r is the radius of rotation, and ω is the angular velocity in radians per second. We can convert the number of revolutions per minute to radians per second using the following equation:

ω = (2 * π * n) / 60

where n is the number of revolutions per minute. Solving for n, we get:

n = sqrt(T / (m * r)) * (60 / (2 * π))

Substituting the values, we get:

n = sqrt(82.0 N / (4.00 kg * r)) * (60 / (2 * π))