Question

A turntable that spins at a constant 76.0 rpm takes 3.50 s to reach this angular speed after it is turned on. Find its angular acceleration (in rad/s2), assuming it to be constant, and the number of degrees it turns through while speeding up.

Answer 1

Answer:

$\Delta \theta = 13.928\,rad\,(798.016\,^{\textdegree})$

Explanation:

The angular acceleration of the the turntable is:

$\alpha = \frac{\omega - \omega_{o}}{\Delta t}$

$\alpha = \frac{(76\,\frac{rev}{min} )\cdot (\frac{2\pi\,rad}{1\,rev} )\cdot (\frac{1\,min}{60\,s} )-(0\,\frac{rad}{s} )}{3.50\,s}$

$\alpha \approx 2.274\,\frac{rad}{s^{2}}$

The change in angular position is:

$\Delta \theta = \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\Delta \theta = \frac{1}{2}\cdot (2.274\,\frac{rad}{s^{2}} )\cdot (3.50\,s)^{2}$

$\Delta \theta = 13.928\,rad\,(798.016\,^{\textdegree})$

Answer 2

Answer: 2.27 rad/s², 796°

Explanation:

Given,

time, t = 3.5 s

speed, w = 76 rpm

To start, we have to convert the speed to rad/s

76 rpm * 1/60 s * 2π rad/rev

= 7.96 rad/s

α = Δω / Δt

α = 7.96 rad/s / 3.5 s

α = 2.27 rad/s²

Θ = 1/2αt²

Θ = 1/2 * 2.27 * 3.5 * 3.5

Θ = 1/2 * 2.27 * 12.25

Θ = 1/2 * 27.8075

Θ = 13.90 rads

Converting back, we have,

Θ = (13.9 * 360°) / 2π rads

Θ = 796°

Therefore the angular acceleration is 2.27 rad/s²

The angle turned through is 13.9 rads or 796°