A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 and has a constant angular acceleration of 2.1 rad/s^2. At what value of t will the radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude?

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

[tex]\omega = \alpha t = 2.1 t[/tex]

And so the radial acceleration is

[tex]a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2[/tex]

The tangential acceleration is always the same since angular acceleration is constant:

[tex]a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2[/tex]

For these 2 quantities to be the same

[tex]a_r = a_t[/tex]

[tex]0.441 t^2 = 0.21[/tex]

[tex]t^2 = 0.21/0.441 = 0.4762[/tex]

[tex]t = \sqrt{0.4762} = 0.69 s[/tex]

AFOKE88 AFOKE88 · Answer 2 · 2022-01-24T15:15:10+01:00

The value of t whereby the radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude is; t = 0.69 s

Calculating Radial and Tangential Acceleration

Let the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude be denoted as t. Thus, angular velocity at time (t) is;

ω = αt

where;

α is angular acceleration

We are given;

radius; r = 10 cm = 0.1 m

constant angular acceleration; α = 2.1 rad/s²

Thus, ω = 2.1t

Now, we can find the radial acceleration from the formula;

α_r = ω²r

Thus;

α_r = (2.1t)² × 0.1

α_r = 0.441 t² m/s²

The tangential acceleration is gotten from the formula is;

α_t = α × r

α_t = 2.1 × 0.1

α_t = 0.21 m/s²

The condition in the question implies that the tangential acceleration is equal to the radial acceleration. Thus;

α_t = α_r

0.21 = 0.441 t²

t = √(0.21/0.441)

t = 0.69 s

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