A 300g rock tied to a string is rotated in a vertical circle (up and down, so gr

Question

A 300g rock tied to a string is rotated in a vertical circle (up and down, so gravity is mportant) with radius 0.5m. Model the rock as a point particle. If the rock starts from rest and a torque of 2Nm is applied for 3s, what is the rotation rate of the rock? What is the tension in the string at the top of the circle? A 300g rock tied to a string is rotated in a vertical circle (up and down, so gravity is mportant) with radius 0.5m. Model the rock as a point particle. If the rock starts from rest and a torque of 2Nm is applied for 3s, what is the rotation rate of the rock? What is the tension in the string at the top of the circle?

Explanation / Answer

Moment of inertia of the rock about the center of the circle is,

I = Mr2

Torque acting on it is T = 2 N-m for t = 3s.

So angular impulse imparted is,

J = Tt = (2Nm)(3s) = 6 Nms

J = L2 - L1, where L2and L1 are the final and initial angular momentum of the rock.

or, J = mvr

or, 6 Nms = (0.3 kg)v(0.5 m)

or, v = 40 m/s, so the rock rotates at the rate of 40 m/s.

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At the top of the circle let the speed be, v, then conservation of energy says,

(1/2)m(40 m/s)2 = mg(2r) + (1/2)mv2

or, (1/2)(40 m/s)2 = g(2r) + (1/2)v2

or, (40 m/s)2 = 4(9.8 m/s2)(0.5 m) + v2

or, v = 39.75 m/s

Let the tension at top be T, then

T = mg + mv2/r = (0.3 kg)[9.8 m/s2 + (39.75 m/s)2/(0.5m)]

or, T = 951.2 N

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

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