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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.

User Nrofis
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5.0k points

2 Answers

4 votes

Answer: 2.27 rad/s², 796°

Step-by-step 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°

User Brijesh Shiroya
by
5.4k points
3 votes

Answer:


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

Step-by-step explanation:

The angular acceleration of the the turntable is:


\alpha = (\omega - \omega_(o))/(\Delta t)


\alpha = ((76\,(rev)/(min) )\cdot ((2\pi\,rad)/(1\,rev) )\cdot ((1\,min)/(60\,s) )-(0\,(rad)/(s) ))/(3.50\,s)


\alpha \approx 2.274\,(rad)/(s^(2))

The change in angular position is:


\Delta \theta = (1)/(2)\cdot \alpha \cdot t^(2)


\Delta \theta = (1)/(2)\cdot (2.274\,(rad)/(s^(2)) )\cdot (3.50\,s)^(2)


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

User Roey Angel
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4.7k points