4 initial angular spued of the skater is 𝜔1 = 3 rotations/rec.
let 𝐼1 be the initial rotational inertia of the skater.
given that final rotational inertia of the skater is
𝐼2 =
𝐼1
2
let 𝜔2 bue the final angular spued of the skatur. From angular momentum conservation
initial angular momentum = final angular momentum
𝐼1 𝜔1
𝐼1 𝑊1
⇒ 𝑊2
𝜔2
𝑊2
= 𝐼2 𝜔2
𝐼1
= 𝜔2
2
= 2𝑊1
=2×3
= 6 rotations/rec.
The linear speed is,
𝑣
2𝜋𝑅
𝑇
2𝜋(8)
=
10
= 5.0265 m/s
= 5.0 m/s( two significant figures )
=
The angular speed is,
𝜔
𝑣
𝑅
2𝜋𝑅
= 𝑇
𝑅
2𝜋
=
𝑇
2𝜋
=
10
= 0.6283rad/s
= 0.63rad/s( two significant figures )
=
acceleration while moving around a circular track = v^2 / r
it does not depend on the mass
so the acceleration will be a) equal
so option a) equal is the correct answer
The torque is calculated as folloes:
Torque = mg [r / 2 ] = [60 kg * 10 m/s2 * 3 m] /2 = 900 N.m
Given that 1
(1)
⇒ 𝐵𝑦 relation we know that.
linear " velocity = radius (𝑟) × angular
(v) welocity (w) distance of particle.

For the particle located at centre of circle 𝑣 centre = 𝑟 × 𝜔 = 0 × 2
ventre = om /s]

For the particle located at the rim of circle 𝑉𝑟𝑖𝑚 = 𝑟 × 𝜔 = 1 m × 2rad/s
𝑉rim = 2 m/s
Thank You.
Given
Radius of circle 𝑅 = 2 m
Rate at which it us acrelerates.
𝑣 = 10 m/s2
Angular arceleration of the
object is given by
𝛼=
𝑉
𝑅
substituting the values
= 10/2
= 5 repes radians /s2
The distance traveled by the object for a complete revolution is equal to the
circemtenence of the cilcle.
𝑑 = 2𝜋𝑅
2𝑚 = 2𝜋𝑅
2𝑚
𝑅=
= 0.318309886 m
2𝜋
𝑅 = 0.3183 m