A mass m = 4.5 kg hangs on the end of a massless rope L = 2.01 m long. The pendu

Question

A mass m = 4.5 kg hangs on the end of a massless rope L = 2.01 m long. The pendulum is held horizontal and released from rest.

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 = 389 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.)

4)Now a peg is placed 4/5 of the way down the pendulum

Explanation / Answer

1) 0.5*m*v^2 = m*g*L

v = sqrt(2*g*L)

v = 6.2766 m/s

2)

T = mg + mv^2/r = m*(9.8 + 6.2766^2/2.01) = 87.98 N

3)

m = T/(g+v^2/r) = 389/(9.8 + 6.2766^2/2.01) = 19.896 kg

4)

0.5*m*v^2 = Change in PE = m*g*L - m*g*2L/5 = 3*m*g*L/5

L = 2.01

v = sqrt(6*g*L/5) = 4.86185 m/s

5) T = mv^2/r - mg = m*(v^2/r - g) = 4.5*(4.86185^2/r - 9.8)

r = L/5 = 2.01/5 = 0.4 m

T = 220.5 N

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