The rock has a mass of 0.6 kilograms and the string is 1.14 meters long. It is spinning at 1.2 radians per second.

Question

The rock has a mass of 0.6 kilograms and the string is 1.14 meters long. It is spinning at 1.2 radians per second.

What is the force Jim must apply on the string to keep the rock spinning at that speed without accelerating?

Answer 1

If velocity is constant during circular velocity

Tangential acceleration at = 0

Tangential acceleration is the rate of change of velocity, velocity is constant therefore no rate of change of velocity

If the rock is spin overhead, draw a force diagram

You have a tension force pointing toward the center and a gravity force point downward.

Sum forces about the centripetal , which is equal to mass time centripetal acceleration

T = m * ac

ac = V^2/R

V = w * R

ac = (w*R)^2/R

ac = w^2 * R

T = m * w^2 * R

T = 0.6 * (1.2)^2 * (1.14

T = 0.984 N

Answer 2

m = 0.6 kg; r = 1.14 m; ω = 1.2 rad/s

Acceleration (ac) towards the center of a circle = v²/r

Tension (T) is the force that keeps the rock moving in a circle, so T = mac = mv²/r

Because v = ωr; T = m(ωr)²/r = mω²r = (0.6 kg)(1.2 /s)²(1.14 m) = 1 kgm/s² = 1 N