College Physics (Instructor's)
College Physics (Instructor's)
11th Edition
ISBN: 9781305965317
Author: SERWAY
Publisher: CENGAGE L
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Chapter 8, Problem 84AP

A light rod of length 2L is free to rotate in a vertical plane about a frictionless pivot through its center. A particle of mass m1 is attached at one end of the rod, and a mass m2 is at the opposite end, where m1 > m2. The system is released from rest in the vertical position shown in Figure P8.84a, and at some later time, the system is rotating in the Position shown in Figure P8.84b. Take the reference point of the gravitational potential energy to be at the pivot, (a) Find an expression for the system's total mechanical energy in the vertical position. (b) Find an expression for the total mechanical energy in the rotated position shown in Figure P8.84b. (c) Using the fact that the mechanical energy of the system is conserved, how would you determine the angular speed co of the system in the rotated position? (d) Find the magnitude of the torque on the system in the vertical position and in the routed position. Is the torque constant? Explain what these results imply regarding the angular momentum of the system, (c) Find an expression for the magnitude of the angular acceleration of the system in the rotated position. Does your result make sense when the rod is horizontal? When it is vertical? Explain.

Chapter 8, Problem 84AP, A light rod of length 2L is free to rotate in a vertical plane about a frictionless pivot through

Figure P8.84

(a)

Expert Solution
Check Mark
To determine
The total mechanical energy of the system about vertical position.

Answer to Problem 84AP

The total mechanical energy of the system about vertical position is m1gLm2gL

Explanation of Solution

Given Info:

The mass of the particles are m1 and m2 and light rod of length 2L . Mass m1 is attached at one end and m2 attached at other end.

In Figure P8.84 (a) the system has only potential energy and at centre the potential energy is zero since the reference point of the gravitational potential energy is at the pivot that is at the midpoint.

The formula to calculate total mechanical energy of the system about vertical position is

E=m1gLm2gL

  • g is the acceleration due to gravity
  • L is the distance of the masses from the pivot

Thus, the total mechanical energy of the system about vertical position is m1gLm2gL

Conclusion:

Therefore the total mechanical energy of the system about vertical position is m1gLm2gL

(b)

Expert Solution
Check Mark
To determine
Total mechanical energy in the rotated position.

Answer to Problem 84AP

The total mechanical energy in the rotated position is 12(m1+m2)L2ω2+(m1m2)gLsinθ

Explanation of Solution

Given Info: The mass of the particles are m1 and m2 and light rod of length 2L . Mass m1 is attached at one end and m2 attached at other end.

In Figure P8.84 (b) the system has potential energy and rotational kinetic energy.

The expression to calculate total mechanical energy of the rotating system is,

E=KE+PE

  • KE is the rotational kinetic energy
  • PE is the new potential energy

The potential energy of the system is given by

PE=m1gLsinθm2gLsinθ (I)

The formula to calculate the rotational kinetic energy is given by

KE=12(I1+I2)ω2

  • I1 is moment of inertia of mass m1
  • I2 is the moment of inertia of mass m2
  • ω is the angular speed of the rotated system

The formula to calculate moment of inertia of mass m1 is

I1=m1L2

The formula to calculate moment of inertia of mass m2 is

I2=m2L2

Rewrite the above equation for KE using the expression for I1 and I2

KE=12(m1L2+m2L2)ω2=12(m1+m2)L2ω2 (II)

Formula to calculate the total mechanical energy is,

E=PE+KE

  • E is the total mechanical energy at the rotated position

Substitute equation (I) and (II) in above equation to calculate E .

E=12(m1+m2)L2ω2+m1gLsinθm2gLsinθ=12(m1+m2)L2ω2+(m1m2)gLsinθ

Conclusion: The total mechanical energy in the rotated position is 12(m1+m2)L2ω2+(m1m2)gLsinθ

(c)

Expert Solution
Check Mark
To determine
The angular speed of the rotated system

Answer to Problem 84AP

The angular speed of the rotated system is  (2g(m1m2)(1sinθ)(m1+m2)L)12

Explanation of Solution

Given Info:

According to conservation of total mechanical energy of the system, the total mechanical energy of the vertical system is same as total mechanical energy of the rotated system.

The formula to calculate total mechanical energy of the system about vertical position is

E=m1gLm2gL

Formula to calculate the mechanical energy of the rotated system is,

E=12(m1+m2)L2ω2+(m1m2)gLsinθ

Equate the above expressions for E and E to calculate ω

12(m1+m2)L2ω2+(m1m2)gLsinθ=m1gLm2gL12(m1+m2)L2ω2=(m1m2)gL(1sinθ)ω=(2g(m1m2)(1sinθ)(m1+m2)L)12

Conclusion:

Therefore the angular speed of the rotated system is  (2g(m1m2)(1sinθ)(m1+m2)L)12

(d)

Expert Solution
Check Mark
To determine
The magnitude of torque in the vertical and rotated position

Answer to Problem 84AP

Therefore the net torque on the vertical system is zero and rotated system is (m1m2)gLcosθ and torque is not a constant therefore angular momentum will change with a non-uniform rate

Explanation of Solution

Given Info:

In the case of vertical position the gravitational force and the position of the masses from the axis of rotation lies in the same line so the net torque on the system is zero.

Formula to calculate the torque on the mass m1 is,

τ1=m1gLsin(π2θ)=m1gLcosθ

Formula to calculate the torque on the mass m2 is,

τ2=m2gLsin(π2θ)=m2gLcosθ

Formula to calculate net torque on the system is,

τ=τ1+τ2

Substitute m1gLcosθ for τ1 and m2gLcosθ for τ2 to calculate τ

τ=m1gLcosθm2gLcosθ=(m1m2)gLcosθ

Conclusion:

Therefore the net torque on the vertical system is zero and rotated system is (m1m2)gLcosθ

Torque is not a constant therefore angular momentum will change with a non-uniform rate.

(e)

Expert Solution
Check Mark
To determine
The magnitude of angular acceleration of the system in the rotated position.

Answer to Problem 84AP

The angular acceleration of the rotated system is ((m1m2)gcosθ(m1+m2)L)

Angular acceleration has maximum value when rod is horizontal and has minimum value when rod is vertical.

Explanation of Solution

Explanation:

Given Info:

Formula to calculate the angular acceleration of the rotated system is given by

α=τ(I1+I2)

Substitute (m1m2)gLcosθ for τ and (m1L2+m2L2) for I1+I2 in the above equation to calculate α

α=(m1m2)gLcosθ(m1L2+m2L2)=(m1m2)gcosθ(m1+m2)L

Thus the angular acceleration of the rotated system is ((m1m2)gcosθ(m1+m2)L)

Conclusion:

When rod is horizontal that is θ is zero

Substitute 0 for θ in the above equation to calculate angular acceleration to calculate angular acceleration

α=((m1m2)gcos0(m1+m2)L)=((m1m2)g(m1+m2)L)

Angular acceleration of the rod when the rod is horizontal is ((m1m2)g(m1+m2)L)

When rod is vertical that is θ is 90.

Substitute 0 for θ in the above equation to calculate angular acceleration to calculate angular acceleration

α=((m1m2)gcos90(m1+m2)L)=0

Angular acceleration of the rod when the rod is vertical is 0. Therefore the rod has maximum angular acceleration when rod is horizontal. This is position corresponds to unstable equilibrium position.

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Chapter 8 Solutions

College Physics (Instructor's)

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