- #1
Peter G.
- 442
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Hi
A horizontal disc has a hole through its center. A string passes through the hole and connects a mass m on top of the disc to a bigger mass M that hangs below the disc. Initially, the smaller mass is rotating on the disc in a circle of radius r. What must the speed of m be such that the big mass stands still.
So, I managed to get to the answer with not too many problems:
Mg = ma
Mg = mv2 / r
and rearranged to: v = √ (Mgr / m)
My problem is, I don't understand what is happening. So, why/when would the big mass stop moving? Is it when the tension in the string due to the centripetal force equals the big mass Mg, so that the forces on the big Mass balance?
Thanks,
Peter G.
A horizontal disc has a hole through its center. A string passes through the hole and connects a mass m on top of the disc to a bigger mass M that hangs below the disc. Initially, the smaller mass is rotating on the disc in a circle of radius r. What must the speed of m be such that the big mass stands still.
So, I managed to get to the answer with not too many problems:
Mg = ma
Mg = mv2 / r
and rearranged to: v = √ (Mgr / m)
My problem is, I don't understand what is happening. So, why/when would the big mass stop moving? Is it when the tension in the string due to the centripetal force equals the big mass Mg, so that the forces on the big Mass balance?
Thanks,
Peter G.