Physics, 21.04.2020 00:00 thnguyen0720

A turntable is off and is not spinning. A 0.8 g ant is on the disc and is 9 cm away from the center. The turntable is turned on and 0.8 s later it has an angular speed of 45 rpm. Assume the angular acceleration is constant and determine the following quantities for the ant 0.4 s after the turntable has been turned on. Express all quantities using appropriate mks units.

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Answer from: elijah207

Complete Question

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$\alpha = 5.89 rad/s^2$

$w__{0.4}}= 2.36 \ rad/s$

v= 0.212m/s

a_t= 0.5301 m/s

a_r = 0.499 m/s

a = 0.7279 m/s

$F_{net}=5.823*10^{-4}N$

Explanation:

From the question we are told that

mass of the ant is $m_a = 0.8g = \frac{0.8}{1000} = 0.00018kg$

The distance from the center is $d = 9cm = \frac{9}{100} = 0.09m$

The angular speed is $w = 45rpm = 45 * \frac{2 \pi }{60} = 1.5 \pi$

The time taken to attain angular acceleration of 45rpm t_1 = 0.8s

The time taken is t_2 = 0.4 s

The angular acceleration is mathematically represented as

$\alpha = \frac{w}{t}$

$= \frac{1.5}{0.8}$

$\alpha = 5.89 rad/s^2$

The angular velocity at time t= 0.4s is mathematically represented as

$w__{0.4s}} = \alpha * t_2$ Recall angular acceleration is constant

= 5.89 * 0.4

$w__{0.4}}= 2.36 \ rad/s$

The linear velocity is mathematically represented as

$v = w__{t_2}} * r$

= 2.36 * 0.09

v= 0.212m/s

The tangential acceleration is mathematically represented as

$a_{t} = \alpha * r$

= 5.89 * 0.09

a_t= 0.5301 m/s

The radial acceleration is mathematically represented as

$a_r = \frac{v^2}{r}$

$= \frac{0.212^2}{0.09}$

a_r = 0.499 m/s

The resultant velocity is mathematically represented as

$a = \sqrt{a_t^2 + a_r^2}$

$= \sqrt{0.53^2 + 0.499^2}$

a = 0.7279 m/s

The net force is mathematically represented as

$F_{net} = 0.0008 * 0.7279$

$F_{net}=5.823*10^{-4}N$

A turntable is off and is not spinning. A 0.8 g ant is on the disc and is 9 cm away from the center.

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A turntable is off and is not spinning. A 0.8 g ant is on the disc and is 9 cm away from the center....

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