1) How fast is the mass moving at the bottom of its path? 2) What is the magnitu

Question

1) How fast is the mass moving at the bottom of its path?

2) What is the magnitude of the tension in the string at the bottom of the path?

3) If the maximum tension the string can take without breaking is Tmax = 603 N, what is the maximum mass that can be used? (Assuming that the mass is still released from the horizontal and swings down to its lowest point.)

How fast is the mass moving at the top of its new path (directly above the peg)?

5) Using the original mass of m = 7.4 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)?

Explanation / Answer

1)

mg*L = 1/2 m v^2

v = sqrt(2*g*L) = 6.55 m/s

2)

T - mg = mv^2/L

T = 7.4 ( 9.81 + 6.55^2/2.19) =217.78 N

3) v will remain same.

Tmax= mass_max( g+ v^2/L )

mass_max = 603/( 9.81+ 6.55^2/2.19) = 20.51 kg

4)

mg L = mg*2L/5 + 1/2 m v^2

v = sqrt( 6gL/5) = 5.08 m/s [ at top of its path]

T + mg = mv^2/(L/5)

T= m(5*5.08^2/2.19-9.81) = 364.97 N

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