Answer to Question #246137 in Molecular Physics | Thermodynamics for Kelani

Question

A 44.0-cm diameter disk rotates with a constant angular acceleration of 2.60 rad/s². It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time.

(a) At t = 2.47 s, find the angular speed of the wheel.

 __rad/s

(b) At t = 2.47 s, find the magnitude of the linear velocity and tangential acceleration of P.

linear velocity ___ m/s

tangential acceleration ___ m/s²

(c) At t = 2.47 s, find the position of P (in degrees, with respect to the positive x-axis).

___ ° counterclockwise from the +x-axis

Answer 1

Radius of the disk r=0.22m

Angular acceleration $α=2.6rad/s^2$

Initial angle $θ_i=57.3^°$

Time difference Δ=2.47s

a) The angular speed of the wheel is

$ω_f =ω_i + αt$

Here, $ω_i=0$ since it starts from rest

$ω_f =0rad/s +(2.6rad/s^2)(2.47s)= 6.422 rad/s$

b) The linear velocity of the point P is

$v=rω_f \\ =(0.22m)(6.422rad/s) \\ =1.4128m/s$

The tangential acceleration is

$a_t=r α \\ =(0.22m)(2.6rad/s^2) \\ =0.572m/s^2$

c) The position of the point after 2.47s is

$Δθ= ω_it +\frac{1}{2}αt^2 \\ =0+\frac{1}{2}(2.6rad/s^2)(2.47s)^2 \\ =7.931rad$

In degrees,

$θ_f=θ_i +(7.931rad)(\frac{180^°}{π} rad) \\ =57.3^° + 455^° \\ =513.3^°$

From x-axis it would be

513.3 – 360= 153.3

The angle is $153.3^°$ from x-axis