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 = 2.8 m/s and the tension in the rope is T = 13.4 N.

1)

How long is the rope?

2)

What is the mass?

3)

If the maximum mass that can be used before the rope breaks is mmax = 0.98 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)

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)?

5)

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)?

httpswww.flipitphysics.com/Course/ViewProblem?unititemID-1530456&enrollmentID-171956-; Homework: Conservative Forces and Potential Energy Homework: Conservative Forces and Potential Energy Chegg Study | Guided Solutions and Study Help | Chegg.com Pendulum Pendulum 2 1 2 3 4 5 6 Standard Exercise Pendulum 2 Standard Exercise Loop the Loop Interactive Example IE Block Spring Incline A mass hangs on the end of a massless rope. reaches the bottom of its path it is moving at a speed v = 2.8 m/s and the tension in the rope is The pendulum is held horizontal and released from rest. When the mass 13.4 N 1) How long is the rope? m Submit 2) What is the mass? Submit 3) If the maximum mass that can be used before the rope breaks is mmax = 0.98 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal) N Submit

Explanation / Answer

1)Let's start with conservation of energy.

Gravitational potential energy lost (mgh) = kinetic energy gained (½mv²).

The string starts off horizontal, so the mass ends up a distance (L) lower when the mass is at the bottom of its path.

mgL = ½mv² {m cancels out}
gL = ½v²

I'll take g = 9.8 m/s²
L = v² /2g = ( 2.8² / 2 * 9.8 ) = 0.40 m

2)Now we know L, we can think about centripetal force. {F = mv²/r}

At the bottom of its swing, the tension in the string is made up of the sum of two components:

The reaction to the weight force (mg) + the centripetal force needed to keep the mass moving in a circular path of radius (L)

T = mg + mv²/L = m(g + v²/L)

m = T /(g + v²/L) = 13.4 / (9.8 + 2.8²/0.40) = 0.45 kg

3) Recall from the conservation of energy calculation above that the velocity (v) at the bottom of the path does not depend on (m). When m = 0.98 kg, v will still be 2.8 m/s

T = m(g + v²/L)
T = 0.98(9.8 + 2.8²/0.40)
T 28.81 N

4) The peg height (h) is 4L/5 below the release point = 0.32 m below.

Conservation of energy again. From (1) above:
mgh = ½mv²

v² = 2gh = (2 * 9.8 * 0.32) = 6.272 m²/s²
v = 2.50 m/s

5) Now we have the string in a circular path around the peg of radius ( r = 0.4 - 0.32) = 0.08 m
The string is horizontal, so the tension in the string is providing only the centripetal force this time.

T = mv²/r = (0.45 * 2.5² / 0.08) = 35.15 N

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