Question

A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an angular displacement of 13 rad. What is its displacement θ at a time of t = 5.0 s?

Answer 1

Answer:

$\theta = 52 rads$

Explanation:

The rotational kinetic equation is given by the formula:

$\theta = w_{0} t + 0.5 \alpha t^{2}$..............(1)

The object is starting from rest, angular speed, w = 0 rad/s

At t = 2.5 sec, angular displacement, θ = 13 rad

Inserting these parameters into equation (1)

$13 = (0 * 2.5) + 0.5 \alpha 2.5^{2}\\\alpha = \frac{13}{3.125} \\\alpha = 4.16 rad/s^{2}$

At time, t = 5.0 sec, we substitute the value of $\alpha$ into the kinematic equation in (1)

$\theta = (0*5) + 0.5 *4.16* 5^{2}\\\theta = 52 rad$

Answer 2

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.