A mass m = 4 kg hangs on the end of a massless rope L = 1.91 m long. The pendulu

Question

A mass m = 4 kg hangs on the end of a massless rope L = 1.91 m long. The pendulum is held horizontal and released from rest.

1 What is the magnitude of the tension in the string at the bottom of the path? ____ N

2) If the maximum tension the string can take without breaking is Tmax = 337 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.) ___kg

Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg. As it wraps around the peg and attains its maximum height it ends a distance of 3/5 L below its starting point (or 2/5 L from its lowest point)

3) How fast is the mass moving at the top of its new path (directly above the peg)? ____ m/s

4 Using the original mass of m = 4 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)? N

Explanation / Answer

a)

Tension in the string at Bottom most point

T = m*g + m*(2*g*1.91)/(1.91) = 3*mg = 120 N

b)

Tmax = 337 N

so Mmax = Tmax/(3*g) = 11.23 Kg

c)

0.5*m*v^2 = m*(3*1.91/5)*g

v = sqrt(2*3*1.91/5*g) = 4.78 m/s

d)

Tension in the string at the top

= 4*4.78^2/(1.91/5) - 4*10 = 199.25 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.