Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 10, Problem 77P

(a)

To determine

Find the energy of the pendulum.

(a)

Expert Solution
Check Mark

Answer to Problem 77P

The energy of the pendulum is 6.1mJ.

Explanation of Solution

Write the equation for total mechanical energy of a pendulum.

E=12mω2A2 (I)

Here, m is the mass of the pendulum, ω is the angular frequency, and A is the amplitude of the oscillating pendulum.

Write the equation for angular frequency of the SHM for pendulum oscillation (Refer Equation 10-26a).

ω=gL (II)

Here, g is the gravitational acceleration and L is the length of the pendulum.

Conclusion:

Substitute equation (II) in (I).

E=12m(gL)2A2=mgA22L (III)

Substitute 0.50kg for m, 9.80m/s2 for g, 5.0cm for A, and 1.0m for L in the equation (III).

E=(0.50kg)(9.80m/s2)[(5.0cm)(0.01m1cm)]22(1.0m)=6.125×103J(1mJ103J)=6.125mJ6.1mJ

Therefore, the energy of the pendulum is 6.1mJ.

(b)

To determine

Find the percentage of the pendulum’s energy lost during one cycle.

(b)

Expert Solution
Check Mark

Answer to Problem 77P

The percentage of the pendulum’s energy lost during one cycle is 1.1%.

Explanation of Solution

Write the equation for time period of the pendulum.

T=2πLg (IV)

Here, T is the time period of the oscillating pendulum.

Write the equation for potential energy of an object.

U=m2gh (V)

Here, U is the potential energy of an object, m2 is the mass of an object to drop in a week, and h is the distance of the object to drop in 1 week.

The loss of oscillation time period of cycle for 1 week is.

cycles per week=1wkT (VI)

The percentage of pendulum’s energy loss for 1 week is.

E(%)=(Ucycles/wk)E×100% (VII)

Here, E(%) is the percentage of the pendulum’s energy lost during one cycle.

Conclusion:

Substitute equation (IV) in (VI).

cycles per week=1wk(604,800s1wk)2πgL=604,800s2πgL (VIII)

Substitute the equation (III), (V) and (VIII) in equation (VII).

E(%)=(m2gh604,800s2πgL(mgA22L))×100%=4πm2h(604,800s)mA2L3g×100%

Substitute 0.50kg for m, 2.0kg for m2, 9.80m/s2 for g, 5.0cm for A, 1.0m for L, and h in above equation.

E(%)=4(3.14)(2.0kg)(1.0m)(604,800s)(0.50kg)[(5.0cm)(0.01m1cm)]2(1.0m)39.80m/s2×100%=1.06%1.1%

Therefore, the percentage of the pendulum’s energy lost during one cycle is 1.1%.

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Chapter 10 Solutions

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