Physics, 08.04.2020 02:29 Nayerlin16

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s and a constant angular acceleration of 0.906 rev/s2rev/s2.(a) Compute the angular velocity of the turntable after 0.200 s. (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turn-table at t = 0.200 s? (d) What is the magnitude of the resultant acceleration of a point on the rim at t = 0.200 s?

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Answer from: boxergirl2062

For the electric turntable 0.730 mm in diameter that is rotating with an initial angular velocity of 0.240 rev/s and a constant angular acceleration of 0.906 rev/s², we have:

a) The angular velocity of the turntable after 0.200 s is 2.65 rad/s.

b) The number of revolutions in 0.200 s is 0.066.

c) The tangential speed of a point on the rim of the turntable at a time of 0.200 s is 9.67x10⁻⁴ m/s.

d) The magnitude of the resultant acceleration of a point on the rim at t = 0.200 s is 3.30x10⁻³ m/s².

a) We can calculate the angular velocity with the following equation:

$\omega_{f} = \omega_{i} + \alpha t$ (1)

Where:

$\omega_{f}$ : is the final angular velocity =?

$\omega_{i}$ : is the initial angular velocity = 0.240 rev/s

α: is the angular acceleration = 0.906 rev/s²

t: is the time = 0.200 s

By entering the above values into equation (1), we have:

$\omega_{f} = 0.240 \frac{rev}{s}*\frac{2\pi rad}{1 rev} + 0.906 \frac{rev}{s^{2}}*\frac{2\pi rad}{1 rev}*0.200 s = 2.65 rad/s$

Hence, the angular velocity of the turntable after 0.200 s is 2.65 rad/s.

b) We can find the number of revolutions as follows

$\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha t^{2}$ (2)

Where:

$\theta_{f}$ : is the final displacement =?

$\theta_{i}$ : is the initial displacement = 0

Then, the number of revolutions is (eq 2):

$\theta_{f} = 0 + 0.240 \frac{rev}{s}*0.200 s + \frac{1}{2}0.906 \frac{rev}{s^{2}}*(0.200 s)^{2} = 0.066 rev$

c) The tangential speed is equal to:

$v = \omega_{f}*r$ (3)

Where:

r: is the radius of the turntable = d/2 = 0.730/2 mm = 0.365 mm

Hence, the tangential speed is (eq 3):

$v = 2.65 rad/s*0.365 \cdot 10^{-3} m = 9.67 \cdot 10^{-4} m/s$

d) The magnitude of the resultant acceleration is equal to the sum of the centripetal and tangential acceleration:

$a_{r} = \sqrt{a_{t}^{2} + a_{c}^{2}}$ (4)

Where:

$a_{c}$ : is the centripetal acceleration = $\omega_{f}^{2}*r$

$a_{t}$ : is the tangential acceleration = α*r

So, the resultant acceleration is (eq 4):

$a_{r} = \sqrt{(\alpha*r)^{2} + (\omega_{f}^{2}*r)^{2}}$

$a_{r} = \sqrt{(0.906 \frac{rev}{s^{2}}*\frac{2\pi rad}{1 rev}*0.365 \cdot 10^{-3} m)^{2} + ((2.65 rad/s)^{2}*0.365 \cdot 10^{-3} m)^{2}} = 3.30 \cdot 10^{-3} m/s^{2}$

Therefore, the magnitude of the resultant acceleration is 3.30x10⁻³ m/s².

Learn more about circular motion here:

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I hope it helps you!

Answer from: bossboybaker

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

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

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

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