Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 6, Problem 45P

(a)

To determine

To Find:The acceleration of each mass.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Smaller mass, m=50.0g = 0.05kg .

Larger mass, M=2.00kg

Distance between two smaller masses, dm=6.00cm = 0.06 m

Distance between smaller mass and larger mass, dmM=5.00cm = 0.05 m

Formula used:

Gravitational force between two bodies having mass m and M , d distance away is:

  F=GmMd2

G is the gravitational constant.

Newton’s second law of motion:

  F=ma

Calculation:

Acceleration of smaller mass (m):

  mam=GmMd mM2am=6.67×1011×2.00 ( 0.05 )2m/s2am=5.34×108m/s2

Acceleration of larger mass (m):

  MaM=GmMd mM2aM=6.67×1011×0.05 ( 0.05 )2m/s2aM=1.33×109m/s2

Conclusion:

Thus, mass m accelerates with 5.34×108m/s2 and mass M accelerates with 1.33×109m/s2 .

(b)

To determine

To Find: The time required by each mass to move by 1 mm.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Smaller mass, m=50.0g = 0.05kg .

Larger mass, M=2.00kg

Distance between two smaller masses, dm=6.00cm = 0.06 m

Distance between smaller mass and larger mass, dmM=5.00cm = 0.05 m

Distance moved by each mass, s=1mm = 1.0×103m

mass m accelerates with 5.34×108m/s2

mass M accelerates with 1.33×109m/s2 .

Formula used:

Second equation of motion:

  s=ut+12at2

Where, u is the initial velocity, t is the time and a is the acceleration.

Calculation:

Consider that each mass is initially at rest. Then, u=0 .

For mass m:

  s=ut+12amt21.0×103=0+12×(5.34× 10 8)t2t= 2× 10 3 5.34× 10 8 t=193.53s

For mass M:

  s=ut+12aMt21.0×103=0+12×(1.33× 10 9)t2t= 2× 10 3 1.33× 10 9 t=1226.28s

Conclusion:

Thus, mass mmoves in 193.53s and mass Mmoves in 1226.28s.

(c)

To determine

To Find: The angle by which the rotates in the given time interval.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

Smaller mass, m=50.0g = 0.05kg .

Larger mass, M=2.00kg

Distance between two smaller masses, dm=6.00cm = 0.06 m

Distance between smaller mass and larger mass, dmM=5.00cm = 0.05 m

Distance moved by each mass, s=1mm = 1.0×103m

Mass m accelerates with 5.34×108m/s2.

Mass M accelerates with 1.33×109m/s2.

Mass m moves in 193.53s

Formula used:

Second equation of rotational motion:

  θ=ωot+12αt2

Where, ωo is the initial angular speed, t is the time and α is the angular acceleration.

Angular acceleration is related to linear acceleration as:

  α=ar

Where, a is the linear acceleration and r is the radius.

Calculation:

Initially, the rod is at rest. So, ωo=0 .

  θ=ωot+12αt2θ=0+12×ar×t2θ=12×5.34× 10 80.06×(193.53)2θ=1.67×102radθ=1.67×102×180πdegθ=0.96°

Conclusion:

Thus, the rod rotates by an angle of 1.67×102rad or 0.96° .

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