Question

A phonograph record accelerates from rest to 41.0 rpm in 4.73 s. (a) What is its angular acceleration in rad/s2? rad/s2 (b) How many revolutions does it go through in the process? rev

Answer 1

Answer:

a) 0.91 rad/s²  

b) 1.61 rev

Explanation:

We have:

t = 4.73 s

$\omega_{f} = 41.0 \frac{rev}{min}* \frac{2\pi rad}{1 rev} * \frac{1 min}{60 s} = 4.29 rad/s$  

ω₀ = 0

a) To find the angular acceleration (α) we can use the following equation:

$\omega_{f} = \omega_{0} + \alpha*t$  

$\alpha = \frac{\omega_{f}}{t} = \frac{4.29 rad/s}{4.73 s} = 0.91 rad/s^{2}$

Therefore, the the angular acceleration is 0.91 rad/s².

b) To find the revolutions (θ) we can use the following equation:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta$  

$\theta = \frac{\omega_{f}^{2}}{2\alpha} = \frac{(4.29 rad/s)^{2}}{2*0.91 rad/s^{2}} = 10.11 rad$

Now, we need to convert rad to rev:

$\theta = 10.11 rad* \frac{1 rev}{2\pi rad} = 1.61 rev$

Hence, the number of revolutions is 1.61.

I hope it helps you!            

Answer 2

Answer:

(a) α = 0.908 rad/s²

(b) Ф = 1.62 rev.

Explanation:

(a)

Using,

α = (ω-ω₀)/t................. Equation 1

Where α = angular acceleration, ω = final angular velocity, ω₀ = Initial angular velocity, t = time.

Given: ω₀ = 0 rad/s (from rest), ω = 41 rpm = 41(0.10472) = 4.29352 rad/s, t = 4.73 s

Substitute into equation 1

α = (4.29352-0)/4.73

α = 0.908 rad/s²

(b)

using,

Ф= (ω+ω₀)t/2...........................Equation 2

Where Ф = revolution in rad.

Given: ω₀ = 0 rad/s. ω = 4.29352 rad/s, t = 4.73 s

Substitute into equation 2

Ф = (4.29352+0)4.73/2

Ф = 10.154 rad

Ф = 10.154/2π = 10.154/(2×3.14)

Ф = 10.154/6.28

Ф = 1.62 rev.