PHYSICS:F/SCI.+ENGRS.,V.1
PHYSICS:F/SCI.+ENGRS.,V.1
10th Edition
ISBN: 9781337553575
Author: SERWAY
Publisher: CENGAGE L
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Chapter 15, Problem 48CP

A smaller disk of radius r and mass m is attached rigidly to the face of a second larger disk of radius R and mass M as shown in Figure P15.48. The center of the small disk is located at the edge of the large disk. The large disk is mounted at its center on a frictionless axle. The assembly is rotated through a small angle θ from its equilibrium position and released. (a) Show that the speed of the center of the small disk as it passes through the equilibrium position is

v = 2 [ R g ( 1 cos θ ) ( M / m ) + ( r / R ) 2 + 2 ] 1 / 2

(b) Show that the period of the motion is

v = 2 π [ ( M / 2 m ) + R 2 + m r 2 2 m g R ] 1 / 2

Figure P15.48

Chapter 15, Problem 48CP, A smaller disk of radius r and mass m is attached rigidly to the face of a second larger disk of

(a)

Expert Solution
Check Mark
To determine

The speed of the center of the small disk as it passes through the equilibrium position is v=2[Rg(1cosθ)(M/m)+(r/R)2+2]1/2.

Answer to Problem 48CP

The speed of the center of the small disk as it passes through the equilibrium position is 2[Rg(1cosθ)(M/m)+(r/R)2+2]1/2.

Explanation of Solution

The radius of the smaller disk is r, the mass of the smaller disk is m, the radius of larger disk is R, the mass of the larger disk is M and the angle at which assembly is rotated is θ.

Consider the figure for the given situation.

PHYSICS:F/SCI.+ENGRS.,V.1, Chapter 15, Problem 48CP

Figure (1)

The loss in the potential energy at S is converted into kinetic energy at Q.

Write the expression for the height of the smaller disk from the centre point O.

    h=OBOA

Here, h is the height of the smaller disk from the centre point O.

Substitute R for OB and Rcosθ for OA in above equation to find h.

    h=RRcosθ=R(1cosθ)

Here, R is the radius of larger disk and θ is the angle at which assembly is rotated.

Write the expression for the loss in potential energy.

    Epotential=mgh

Here, Epotential is the loss in potential energy, m is mass of the smaller disk and g is the acceleration due to gravity.

Substitute R(1cosθ) for h in above equation to find Epotential.

    Epotential=mgR(1cosθ)

Write the expression for the moment of inertia of the larger disk about the cylinder axis.

    Ilargerdisk=MR22

Here, Ilargerdisk is the moment of inertia of the larger disk about the cylinder axis, M is mass of the larger disk and R is the radius of larger disk.

Write the expression for the moment of inertia of the smaller disk about the cylinder axis.

    Ismalldisk=mr22

Here, Ismalldisk is the moment of inertia of the smaller disk about the cylinder axis and r is the radius of smaller disk.

Write the expression for the moment of inertia of the smaller disk about the diameter.

    I=mR2

Here, I is the moment of inertia of the smaller disk about the diameter.

Write the expression for the net moment of inertia of the two disk system.

    I=Ilargerdisk+Ismalldisk+I

Here, I is the net moment of inertia of the two disk system.

Substitute MR22 for Ilargerdisk, mr22 for Ismalldisk and mR2 for I in the above equation to find I.

    I=MR22+mr22+mR2

Write the expression for the angular velocity of the disk.

    ω=vR

Here, ω is the angular velocity of the disk and v is the linear velocity of the disk.

The gain in kinetic energy of the system is equal to the sum of the center of mass of the small disk, the rotational energy of the larger disk and the rotational energy of the smaller disk about O.

 Write the expression for the gain in kinetic energy of the system.

    Ekinetic=12Iω2

Here, Ekinetic is the gain in kinetic energy of the system.

Substitute (MR22+mr22+mR2) for I and vR for ω in the above equation to find Ekinetic.

    Ekinetic=12(MR22+mr22+mR2)(vR)2

Apply conservation law of energy.

    Epotential=Ekinetic

Substitute 12(MR22+mr22+mR2)(vR)2 for Ekinetic and mgR(1cosθ) for Epotential in the above equation.

    mgR(1cosθ)=12(MR22+mr22+mR2)(vR)2(vR)2=2mgR(1cosθ)(MR22+mr22+mR2)v2=2mgR(1cosθ)(MR22+mr22+mR2)R2v=[2mgR(1cosθ)(MR22+mr22+mR2)]1/2R

Further solve the above equation.

    v=[2gR(1cosθ)R22(Mm+mr2mR2+mm)]1/2R=[4gR(1cosθ)(M/m)+(r/R)2+2]1/2=2[Rg(1cosθ)(M/m)+(r/R)2+2]1/2

Conclusion:

Therefore, the speed of the center of the small disk as it passes through the equilibrium position is 2[Rg(1cosθ)(M/m)+(r/R)2+2]1/2.

(b)

Expert Solution
Check Mark
To determine

The period of the motion is T=2π[(M+2m)R2+mr22mgR]1/2.

Answer to Problem 48CP

The period of the motion is 2π[(M+2m)R2+mr22mgR]1/2.

Explanation of Solution

As the value of angle at which assembly is rotated is very small.

    sinθθ

From the figure, write the expression for the equation of motion.

    Id2θdt2=mgRsinθ

Substitute θ for sinθ in above equation.

    Id2θdt2=mgRθd2θdt2+mgRIθ=0        (1)

Write the expression for the equation of motion.

    d2θdt2+ω2θ=0        (2)

Compare equations (1) and (2).

    ω2=mgRIω=mgRI

Formula to calculate the period of the motion is,

    ω=2πTT=2πω

Here, T is the period of the motion.

Substitute mgRI for ω in the above equation.

    T=2πmgRI

Substitute (MR22+mr22+mR2) for I in the above equation.

    T=2πmgR(MR22+mr22+mR2)=2π[MR2+mr2+2mR22mgR]1/2=2π[(M+2m)R2+mr22mgR]1/2

Conclusion:

Therefore, the period of the motion is 2π[(M+2m)R2+mr22mgR]1/2.

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Chapter 15 Solutions

PHYSICS:F/SCI.+ENGRS.,V.1

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