A ball of mass M attached to a string of length L moves in a vertical plane counterclockwise.At the top of the circular path, the tension in the string is twice the weight of the ball....

Question

At the bottom, the ball just clears the ground. Air resistance is negligible. Express all answers in terms of M, L and g.

(a) Determine the magnitude and direction of the net force on the ball when it is at the top. (3 Mg, downward)

(b)Determine the speed v0 of the ball at the top. (sqrt(3gL))

The string is cut when the ball is at the top of its circular path.

(c) Determine the time it takes the ball to reach the ground. (2sqrt(L/g))

(d)Determine the horizontal distance the ball travels before hitting the ground (2sqrt(3L))

The answers are in parenthesis I just don't know how to get there.

Answer 1

(a)    Draw a free-body diagram. There are two forces acting on the ball at this point. Tension and force of gravity, mg. Both are pointing downward. You are given T = 2 mg --> F = Mg + 2Mg = 3Mg downward

(b)   Newton’s second law says F = ma, and the centripetal acceleration is a = v^2 / r  , r = L the length of the string --> 3Mg = M(v^2)/L Solve for v.

(c)    The ball is moving in a circular orbit of radius L, at the top of the circle it is 2L above the ground, since the “ball just clears the ground” at the bottom of the arc. You have the horizontal velocity from (b), there is no velocity in the vertical since the tangent to the circle is horizontal at the top. 2L = (1/2)(3g) t^2. Solve for t

(d)   Once you have t, then solve for the horizontal distance x = v t, where v is the value from b and t is the value from c

You should be able to solve from here

Answer 2

Hi Sarah, I have solved this problem sometime back!! here is the solution reposted.

see: https://www.wyzant.com/resources/answers/729716/circular-motion-question

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a) If you draw the FBD of the problem, you easily see that the net force, F_net = Mg + 2(w); where w = mg= weight ==> F_net = Mg + 2Mg = 3Mg, downward direction!

b) This a circular motion problem, this means that F_net is also a centripetal force

==> F_net = mv²/L ==> 3Mg = Mv²/L, solving for the speed, v yields

v = √(3Lg)

c) using the kinematic equation x = v₀t +.5at², but v₀ = 0 and a = g, x = 2L

==> t = √(2x/g) or t = 2√(L/g)

d) to find the horizontal distance, use the formula: x = vt [ remember that v and t are already calculated above]