2. [Ipt A ball of mass m is attached to a string of length L. It is swung in a v
ID: 1783157 • Letter: 2
Question
2. [Ipt A ball of mass m is attached to a string of length L. It is swung in a vertical circle with enough speed to keep the string taut throughout the motion. Assume the ball travels freely in the circle with negligible loss of mechanical energy. Determine if the following six statements are true or false; e.g., enter TTFFFF i. The speed of the ball is constant during the motion. ii. mv,31 = Tt mg, where v, and T, are the speed of the ball and the tension of the string at the top of the circle. iii, The acceleration of the ball at the bottom of the circle is directed 'up' iv. The tension in the string is the same at the top of the circle and at the bottom of the circle V. The tension in the string at the top of the circle is always greater than the weight of the ball vi. The tensions at the top and at the bottom of the circle depend on the average speed of the ball Answer: SumA All Answers 3.[lpt Assume that m-0.40 kg and L = 1.33 m What is the difference in tension, -T, between when the ball is at the bottom and when t sat the top Answer: Subm a AI AnsworExplanation / Answer
i. if there is no loss of mechanical energy then speed will change as vertical position of ball change.
False
ii. at top of circle,
T - downward and mg - donward
and a - centriperal acc that will be downward
hence T + m g = m v^2 / L
False
iii. True. (acceleration is towards the center of circle)
iv. No tension will change for every positon.
False
v. at top it will be less.
vi. False
FFTFFF
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3. Tt = 0
and at top, m g = m v^2 /L => v^2 = L g
Applying energy conservation to find speed at bottom,
m v^2 / 2 + m g (2L) = 0 + m vb^2 /2
g L/2 + 2 g L = m vb^2 /2
vb^2 = 5 g L
at bottom,
tb = mg + m vb^2 / L = m g + 5 m g
= 6 m g
Ans: 6 x .40 x9.8 = 23.52 N
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