A .700 kg ball is on the end of a rope that is 1.30 m in length. The rope is att

Question

A .700 kg ball is on the end of a rope that is 1.30 m in length. The rope is attached to a pole and the entire system (pole, ball, and rope) rotates. The rope makes an angle with the pole. In the figure, the arrow shows the sense of the rotation for the system.

(b.) Apply Newton's second law to the ball. Your answer can only include the forces from part a, the angle  , the mass of the ball m, and the radial acceleration of the ball m, and the radial acceleration of the ball a_r.

(c.) If the angle  =50 degrees, (i) what is the speed of the mass and (ii) the period of the period of the motion?

Explanation / Answer

b) m = 0.700 kg, L = 1.30 m

2 forces act on the ball: gravity mg and tension T

Horizontal direction: Tsin = mv^2/r

where r = Lsin

vertical direction: Tcos = mg

c) So T = mg/cos (from 2nd equation)

Substituting this in the 1st equation, we get

mgsin/cos = mv^2/(Lsin)

(i) v = sqrt(gL(sin)^2/cos)

= sin*sqrt(gL/cos)

= sin(50.0°)* sqrt(9.81*1.30/cos(50.0°))

= 3.41 m/s

(ii) Period T = 2pi*r/v = 2pi*Lsin/v

= (2pi*1.30*sin50)/3.41 = 1.83s

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