Aturntable that spins at a constant 78.0 rpm takes 4.00 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. you may want to review (page) . for related problem-solving tips and strategies, you may want to view a video tutor solution of rotation of a bicycle wheel. part apart complete express your answer in radians per second squared. αα = 2.04 rad/s2 submitprevious answers correct set up: ω0=0. ω=(78.0rpm)(2πrad1rev)(1min60s)=8.1 7rad/s solve: ω=ω0+αt gives α=ω−ω0t=8.17rad/s−04.00s=2.04rad/s2 significant figures feedback: your answer 2.042 rad/s2 was either rounded differently or used a different number of significant figures than required for this part. part b express your answer in degrees.

a) 2.04 rad/sec² b) 936°

Explanation:

a) By definition, angular acceleration, is the rate of  change  of angular velocity, so we can express it as follows:

γ = (ωf -ω₀) / t

The turntable starts from rest, so ω₀ = 0

Now, as the answer is needed to be expressed in rad/sec², first of all, we need to convert ωf from rpm to rad/sec, as follows:

ωf = 78 rev/min. (1 min/60 sec) . ( 2π rad/ rev) = 13/5 π rad/sec

∴ γ = 13/5π / 4.0 rad/ sec² = 13/20 π rad/sec² = 2.04 rad/sec²

b) As the angular acceleration is assumed constant, we can use the kinematic equation for angular displacement, similar to the one used for linear displacement x, as follows:

θ = θ₀ + ω₀t + 1/2γt²

If we choose  θ₀ = 0°,  and we already know that ω₀ = 0, we finally have:

θ = 1/2 γ t²

Replacing by the values, we have:

θ = 1/2 . 13/20 π rad/sec² . (4.0)² sec² = 26/5 π rad

Converting to degrees, we have:

θ = (26/5) π  rad .(360º/2π rad) = 936º