A mass hangs on the end of a massless rope. The pendulum is held horizontal and

Question

A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.3 m/s and the tension in the rope is T = 22.8 N.

1) How long is the rope? m

2) What is the mass? kg

3) If the maximum mass that can be used before the rope breaks is mmax = 1.77 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.) N

4)

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.

How fast is the mass moving when it is at the same vertical height as the peg (directly to the right of the peg)? m/s

5) What is the tension in the string when the mass is at the same vertical height as the peg (directly to the right of the peg)? N

please show work with formulas used and explain

Explanation / Answer

1) v^2= 2gL

= 2.3^2 = 2*9.81 L

= 5.29 = 19.62L

L = 0.27M

2) LT/(v^2 + Lg) =m

= 0.27*22.8/(2.3^2+0.27*9.81)

M = 0.775kg

3) T = mv^2/L + mg

= 1.77*2.3^2/0.27+1.77*9.81

= 0.530

4) v^2 = 2gh...... h=(4/5)L

2*9.81*4/5*0.27

= 2.05m/s

5) You use the velocity you found in question 4 divided by (1/5) of the radius you found in question 1 to find the acceleration. a = v^2/(1/5)r

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