COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 8, Problem 69QAP
To determine

Minimum height that the marble must start from to make it around the loop.

Expert Solution & Answer
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Answer to Problem 69QAP

  h=0.54m

Explanation of Solution

Given info:

Spherical marble of,

  Radius =r0=0.500cm

  Mass =mo=50.0g

Loop-the-loop track,

  Radius =rl=20.0cm

Spherical marble rolls without slipping down the track.

Marble starts from rest and just barely clears the loop to emerge on the other side of the track.

Formula used:

Let's name the minimum height that the marble must start from to make it around the loop

As h.

Let's name the angular velocity of spherical marble at the top of the track as ω.

Let's name the linear speed of spherical marble at the top of the track as vs

Let's name the moment of inertia of spherical marble as Is.

  g=10ms2.

Conservation of mechanical energy:

  Ki+Ui=Kf+Uf

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

  K=12Mv2s+12Isω2...(1)

Condition for rolling without slipping:

  vs=roω...(2)

Equilibrium of vertical forces of spherical marble at the top of the track,

Centrifugal power = weight

  mo(vs2rl)=mog

Calculation:

Let's consider the motion of sphere,

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

The initial gravitational potential energy is Ui=mogh

Final gravitational potential energy is Uf=mog(rl+rl)=2mogrl

Conservation of mechanical energy:

  Ki+Ui=Kf+Uf

  (0)+mogh=Kf+2mogrl

  Kf=mog(h2rl)...(A)

Let's consider the kinetic energy (Kf) of spherical marble at the top of the track,

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

In terms of vs.

Using (1) expression,

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

  K=12Mv2s+12Isω2...(1)

  Kf=12mov2s+12Isω2

Condition for rolling without slipping:

  vs=roω...(2)

  ω=vsro

Substitute into kinetic energy equation:

  Kf=12mov2c+12Icω2

  Kf=12mov2s+12Is( v s r o )2

  Kf=12(mo+Isr02)vs2

From the general knowledge we know that moment of inertia of a sphere is 25m0ro2.

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

  Kf=12(mo+Isr02)vs2

  Kf=12(mo+( 2 5 m 0 r o 2 )ro2)vs2

  Kf=12(mo+25m0)vs2

  Kf=12(75m0)vs2

  Kf=710movs2...(B)

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

  (A)=(B)

  mog(h2rl)=710movs2

  mogh=710movs2+2mogrl

  gh=710vs2+2grl...(3)

Marble starts from rest and just barely clears the loop to emerge on the other side of the track.

So,

Equilibrium of vertical forces of spherical marble at the top of the track,

Centrifugal power = weight

  mo(vs2rl)=mog

  (vs2rl)=g

  vs2=rlg

Substituting vs2 value to (3) equation,

  gh=710vs2+2grl

  gh=710(rlg)+2grl

  h=710rl+2rl

  h=2710rl

Let's substitute the rl value,

  h=2710(0.20m)

  h=0.54m

Conclusion:

Thus, minimum height that the marble must start from to make it around the loop is 0.54m.

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Chapter 8 Solutions

COLLEGE PHYSICS

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