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 = 22.5 N.
How long is the rope?
What is the mass?
If the maximum mass that can be used before the rope breaks is mmax = 1.74 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)
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)?
Return to the original mass. What is the tension in the string at the same vertical height as the peg (directly to the right of the peg)?

Answer 1

sure, so, what did you try?

Answer 2

i tried the length..but other then distance equals velocity over time im drawing a blank

Answer 3

use energy.

Answer 4

delta PE = delta KE

Answer 5

dont i need to know the mass ?

Answer 6

no.

Answer 7

m is on both sides of the equation.

Answer 8

1/2*m*V^2 = mg*delta h

Answer 9

oo ok i got it ..sorry i just have all these in depth problems and i get so confused cause its so much

Answer 10

you can go back now and use L (and using the tension given) to find mass

Answer 11

do i do tension=mgd?

Answer 12

nope thats not it

Answer 13

yeah, my drawing wasn't great. I made it general... specific to this problem it should be:

Answer 14

because it's initially held horizontal...

Answer 15

Tension at the bottom is given.

Answer 16

what does Fc need to be? we know V and r ...

Answer 17

vsquared over r?

Answer 18

mV^2/r yes

Answer 19

so now you have r and have been given V and T... solve for m?