Question

A 2.30kg hoop 2.20m in diameter is rolling to the right without slipping on a horizontal floor at a steady 3.00rad/s .Part AHow fast is its center moving?Part BWhat is the total kinetic energy of the hoop?Part CFind the magnitude of the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop, (iii) a point on the right side of the hoop, midway between the top and the bottom.Separate your answers by commas.Part EFind the magnitude of the velocity vector for each of the points in part (C), except as viewed by someone moving along with same velocity as the hoop.Separate your answers by commas.

Answer 1

A)

`3.30` ms⁻¹

B)

`25.05` J

C)

i)

`6.60` ms⁻¹

ii)

`0` ms⁻¹

iii)

`4.67` ms⁻¹

Step-by-step explanation:

A)

`v_(cm)` = velocity of center of hoop

w = angular speed of the hoop = 3 rads⁻¹

d = diameter of the hoop = 2.20 m

radius of the hoop is given as

`r =(d)/(2) =(2.20)/(2) = 1.10 m`

velocity of center of hoop is related to angular velocity by the relation given as

`v_(cm) = rw`

inserting the values

`v_(cm) = (1.10) (3)`

`v_(cm) = 3.30` ms⁻¹

B)

`I` = moment of inertia of hoop

`m` = mass of the hoop = 2.30 kg

`r` = radius of hoop = 1.10 m

Moment of inertia of the hoop is given as

`I = m r^(2)`

`I = (2.30) (1.10)^(2) = 2.783` kgm²

Total kinetic energy is given as

`T = (0.5) m v_(cm)^(2) + (0.5) I w^(2)`

`T = (0.5) (2.30) (3.30)^(2) + (0.5) (2.783) (3)^(2)\\T = 25.05` J

C)

i)

`v_(h)` = velocity at highest point

velocity at highest point is given as

`v_(h) = 2 v_(cm)`

`v_(h) = 2 (3.30) \\v_(h) = 6.60` ms⁻¹

ii)

`v_(b)` = velocity at bottom

velocity at bottom point is given as

`v_(b) = v_(cm) - v_(cm)`

`v_(h) = 3.30 - 3.30 \\v_(h) = 0` ms⁻¹

iii)

`v_(r)` = velocity at right

velocity at right is given as

`v_(r) = √(2) v_(cm) \\v_(r) = √(2) (3.30)\\v_(r) = 4.67` ms⁻¹