What are the three components of the equation for the second law of motion?

A disk rotates at constant angular acceleration, from angular position θ1 = 16.0 rad to angular position θ2 = 76.0 rad in 5.30 s

Oliga [24]

Answer:

(a) the angular velocity at θ1 is 11.64 rad/s

(b) the angular acceleration is 0.12 rad/ $s^{2}$

(c) the angular position was the disk initially at rest is - 428.27 rad

Explanation:

Given information :

θ1 = 16 rad

θ2 = 76 rad

ω2 = 11 rad/s

t = 5.3 s

(a) The angular velocity at θ1

First, we use the angular motion equation for constant acceleration

Δθ = (ω1+ω2)t/2

θ2 - θ1 = (ω1+ω2)t/2

ω1 + ω2 = 2 (θ2 - θ1) / t

ω1 = (2 (θ2 - θ1) / t ) - ω2

= (2 (76-16) / 5.3) - 11

= 11.64 rad/s

(b) the angular acceleration

ω2 = ω1 + α t

α t = ω2 - ω1

α = (ω2 - ω1)/t

= (11.64 - 11) / 5.3

= 0.12 rad/ $s^{2}$

(c) the angular position was the disk initially at rest, θ0

at rest ω0 = 0

ω2^2 = ω01 t + 2 α Δθ

2 α Δθ = ω2^2

θ2 - θ0 = ω2^2 / 2 α

θ0 = θ2 - (ω2^2) / 2 α

= 76 - ( $11^{2}$ / 2 x 0.12

= 76 - 504.16

= - 428.27 rad

4 0

3 years ago

A ball is thrown vertically upward from the edge of a bridge 22.0 m high with an initial speed of 16.0 m/s. The ball falls all t

KonstantinChe [14]

To solve this problem we will apply the concepts related to the kinematic equations of motion. We will start calculating the maximum height with the given speed, and once the total height of fall is obtained, we will proceed to calculate with the same formula and the new height, the speed of fall.

The expression to find the change in velocity and the height is,

$v_f^2-v_0^2 = -2gh$