Question

Part b.
A thin hoop of mass 22 kg and radius 1.5 m rolls down an incline that is 4 meters tall. What is the velocity of the thin hoop at the bottom of the incline?

m/s



Part c.
A solid disk of mass 22 kg and radius 1.5 m rolls down an incline that is 4 meters tall. What is the velocity of the solid disk at the bottom of the incline?

m/s

5.

[–/1 Points]DETAILSMY NOTES

A 10.3 kg tire that has a moment of inertia of 2340 kgm^2 is initially turning at 20.5 revolutions/s. It coasts to rest after 131 revolutions. What is the magnitude of the torque that slowed it?

N-m

6.

[–/1 Points]DETAILSMY NOTES

With her arms held against her body, a figure skater has a moment of inertia of 0.65 kg-m^2. With her arms extended, her rotational speed is 0.21 rev/s. But when she draws in her arms, her rotational speed is 0.74 rev/s. What is her moment of inertia when her arms are extended?

kg-m^2

Answer 1

(b) Mass of the hoop = m = 22 kg, and its radius = R = 1.5 m.

Initial height of the hoop = h = 4 m.

Hence, initial total energy = P = Potential energy at a height h = mgh,

where, g = 9.8 m / s² is the acceleration due to gravity.

If the angular velocity of the hoop at the bottom is in rad / s,

then its rotational kinetic energy at the bottom of the incline is : K = I² / 2,

where, I = mR² is the moment of inertia of the hoop.

From principle of conservation of angular momentum, we get :

P = K

or, mgh = I² / 2

or, = ( 2mgh / I ) = ( 2mgh / mR² ) = ( 2gh / R² )

or, = ( 2 x 9.8 x 4 / 1.5² ) rad / s = 5.9 rad / s.

Hence, velocity of the thin hoop at the bottom of the incline = v = R = ( 5.9 x 1.5 ) m / s = 8.85 m / s.

(c) Mass of the disk = m = 22 kg, and its radius = R = 1.5 m.

Initial height of the disk = h = 4 m.

Hence, initial total energy = P = Potential energy at a height h = mgh,

where, g = 9.8 m / s² is the acceleration due to gravity.

If the angular velocity of the hoop at the bottom is in rad / s,

then its rotational kinetic energy at the bottom of the incline is : K = I² / 2,

where, I = mR² / 2 is the moment of inertia of the hoop.

From principle of conservation of angular momentum, we get :

P = K

or, mgh = I² / 2

or, = ( 2mgh / I ) = ( 4mgh / mR² ) = ( 4gh / R² )

or, = ( 4 x 9.8 x 4 / 1.5² ) rad / s = 8.344 rad / s.

Hence, velocity of the solid disk at the bottom of the incline = v = R = ( 8.344 x 1.5 ) m / s ~ 12.5 m / s.