Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 9, Problem 101P

(a)

To determine

The distance moved by the ball s1 and time taken t1 to cover the distance s1 before it begin to roll without slipping, and the final speed v1 of the ball.

(a)

Expert Solution
Check Mark

Answer to Problem 101P

  s1=12v0249μkg

  t1=27v0μkg

  v1=57v0

Explanation of Solution

Given: Mass of the ball is M .

Radius of the ball is R .

The coefficient of kinetic friction between the ball and the floor is μk .

At the instant the ball touches the floor it is moving horizontally with a speed v0 , that is the initial velocity.

Formula used:

Velocity v of the ball at any instant of time t is given by

  v=u+at  (1)

Equation of kinematics

  v2=u2+2as   (2)

Where, v is final velocity

  u is initial velocity

  a is acceleration

  s is distance travelled.

Calculation:

  Physics for Scientists and Engineers, Chapter 9, Problem 101P

FIGURE: 1

In the figure 1 , Mg is the force due to gravity which is acting downward through the center and

N is the normal reaction force which is acting in the upward direction through the point of contact.

Velocity v1 of the ball at any instant of time t1 can be written by using equation (1) as,

  v1=v0+at1   (3)

In this problem there arise a force of kinetic friction, fk=μkN acting horizontally through the point of contact and is in the direction opposite to the motion of the ball. This forces of kinetic friction produces horizontal acceleration

  a=ForceMass=μkNM=μkMgM=μkg

This acceleration is in the direction opposite to the direction of motion of the ball and hence equation (3) becomes

  v1=v0μkgt1  (4)

The horizontal force fk also produces a torque about the center of the ball. This torque has a magnitude of fkR , which is acting in anticlockwise direction and produces anticlockwise angular acceleration α in the ball about its center. This can be written as,

  fkR=Icmα  (5)

Where, Icm is the moment of inertia about the center of mass of the ball

  fk=μkN=μkMg

  Icm=25MR2

Thus, the equation (5) becomes,

  μkMgR=25MR2α

Angular acceleration,

  α=52μkgR  (6)

This angular acceleration sets the ball rotating with angular velocity ω in anticlockwise direction, at any instant of time t , given by,

  ω=αt1  (7)

When the ball starts rolling, the velocity

  v1=ωR  (8)

is satisfied.

Substituting equation (8) in equation (4) ,

  ωR=v0μkgt1

Substituting for ω from equation (7) , the above equation becomes,

  αtR=v0μkgt1

Substituting the expression for α from equation (6) and simplifying,

  52μkgt1=v0μkgt1

  52μkgt1+μkgt1=v0

  v0=72μkgt1  (9)

  t1=27v0μkg

Substituting the value of v0 from equation (9) in equation (4) and simplifying,

  v1=72μkgt1μkgt1=52μkgt1=52μkg(27v0μkg)

  v1=57v0

The distance moved by the ball s1 before it begin to roll without slipping:

Using the equation (2) ,

  v12=v02+2as1

Substituting the expressions for v1 and a ,

  (5v07)2=v022μkgs1

  2μkgs1=(5v07)2v02

  2μkgs1=25v0249v02

  2μkgs1=24v0249

  s1=12v0249μkg

Conclusion:

The distance moved by the ball before it begin to roll without slipping is s1=12v0249μkg .

Time taken by the ball to cover the distance s1 before it begins to roll without slipping is t1=27v0μkg .

The final speed of the ball is v1=57v0 .

(b)

To determine

The ratio of the final kinetic energy to the initial kinetic energy of the ball.

(b)

Expert Solution
Check Mark

Answer to Problem 101P

  KfKi=57

Explanation of Solution

Given: Mass of the ball is M ,

Radius of the ball is R

The coefficient of kinetic friction between the ball and the floor is μk

At the instant the ball touches the floor it is moving horizontally with a speed v0 , that is the initial velocity.

Final velocity of the ball is v1

Formula used:

The final kinetic energy of the ball is given by,

  Kf=Ktrans+Krot

  Kf=12Mv12+12Iω2  (10)

Where, I is moment of inertia.

Initial kinetic energy of the ball is given by,

  Ki=12Mv02  (11)

Calculation:

Moment of inertia of the ball is given by,

  I=25MR2

For rotational motion of the ball the velocity,

  v1=Rωω=v1R

Substituting for I and v1 in equation (10) ,

  Kf=12Mv12+12(25MR2)(v1R)2

  Kf=12Mv12+15Mv12

  Kf=710Mv12

But, v1=57v0

Therefore final kinetic energy of the ball is,

  Kf=710M(57v0)2=514Mv02  (12)

From equations (11) and (12) , the ratio of the final kinetic energy to the initial kinetic energy of the ball is,

  KfKi=514Mv0212Mv02

  KfKi=57

Conclusion:

The ratio of the final kinetic energy to the initial kinetic energy of the ball is 57 .

(c)

To determine

The values of s1,t1 and v1 for v0=8.0 m/s and μk=0.060 .

(c)

Expert Solution
Check Mark

Answer to Problem 101P

  s1=27 m

  t1=3.9 s

  v1=5.7 m/s

Explanation of Solution

Given: v0=8.0 m/s and μk=0.060

Formula used:

  s1=12v0249μkg  (13)

  t1=27v0μkg  (14)

  v1=57v0  (15)

Calculation:

Substituting the numerical values in equation (13) ,

  s1=12(8.0 m/s)249×0.060×9.8 m/s2s1=27m

Substituting the numerical values in equation (14) ,

  t1=278.0 m/s0.060×9.8m/s2t1=3.9s

Substituting the numerical values in equation (15) ,

  v1=57(8.0m/s)=5.7m/s

Conclusion:

The values of s1,t1 and v1 are 27 m, 3.9 s, and 5.7 m/s respectively.

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Chapter 9 Solutions

Physics for Scientists and Engineers

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