COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 8, Problem 67QAP
To determine

Cylinder's speed at the bottom of the ramp.

Expert Solution & Answer
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Answer to Problem 67QAP

  vc=3.36ms1

Explanation of Solution

Given info:

Uniform solid cylinder of,

  Radius =r0=5cm

  Mass =θ=25

  =m0=3.00kg

An inclined plane,

  Length =l0=2.00m

  Angle which tilted with horizontal plane =θ=25

Cylinder rolls without slipping down the ramp.

Formula used:

Let's name the vertical height of the plane as h.

Let's name the angular velocity of cylinder at the bottom of the ramp as ω.

Let's name the linear speed of cylinder at the bottom of the ramp as vc

Let's name the moment of inertia of cylinder as Ic.

  g=10ms2.

Conservation of mechanical energy:

  Ki+Ui=Kf+Uf

Kinetic energy for an object that undergoes both translation and rotation:

  K=12Mv2c+12Icω2...(1)

Condition for rolling without slipping:

  vc=roω...(2)

Calculation:

Let's consider the motion of cylinder,

Initially the cylinder is at rest with zero kinetic energy, so Ki=0.

The initial gravitational potential energy is Ui=mogyi

Final gravitational potential energy is Uf=mogyf

Conservation of mechanical energy:

  Ki+Ui=Kf+Uf

  (0)+mogyi=Kf+mogyf

  Kf=mog(yiyf)

But, according to the data given,

  (yiyf)=h

So,

  Kf=mogh...(A)

Let's consider the kinetic energy (Kf) of cylinder at the bottom of the ramp,

Kinetic energy is part translational and part rotational. We can use (2) equation to write ω

In terms of vc.

Using (1) expression,

Kinetic energy for an object that undergoes both translation and rotation:

  K=12Mv2c+12Icω2...(1)

  Kf=12mov2c+12Icω2

Condition for rolling without slipping:

  vc=roω...(2)

  ω=vcro

Substitute into kinetic energy equation:

  Kf=12mov2c+12Icω2

  Kf=12mov2c+12Icω2

  Kf=12mov2c+12Ic( v c r o )2

  Kf=12(mo+Icr02)vc2

From the general knowledge we know that moment of inertia of a cylinder is Ic=12m0ro2.

So, let's substitute the Ic value in to the equation,

  Kf=12(mo+Icr02)vc2

  Kf=12(mo+( 1 2 m 0 r o 2 )ro2)vc2

  Kf=12(mo+12m0)vc2

  Kf=12(32m0)vc2

  Kf=34movc2...(B)

Since (A),(B) equations are equal,

  (A)=(B)

  mogh=34movc2

  gh=34vc2

  vc2=4gh3

  vc=4gh3

Let's substitute the values,

  vc=4*10ms 2*2msin 253

  vc=11.27

  vc3.36ms1

  vc=3.36ms1

Conclusion:

Thus, cylinder's speed at the bottom of the ramp is 3.36ms1.

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Chapter 8 Solutions

COLLEGE PHYSICS

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