Physics, 02.06.2020 18:58 Andjew

Consider a spinning plate is dropped onto a stationary plate (which is itself at rest on a frictionless surface). Both plates have a radius of 30cm and a mass of 1kg. The spinning plate is initially spinning at a rate of 0.7 revolutions per second. Hint: This is like a totally-inelastic collision.

Required:
a. After a sufficiently long time, what is the angular velocity of the initially-spinning plate? What about the initially-stationary plate?
b. Assume that the period of velocity matching happens over a course of 2 seconds. Further, assume that the torque exerted by each plate on the other is constant over time. In that case, what is the magnitude of the acceleration that each plate feels during those two seconds? Hint: Use the rotational impulse-momentum theorem.

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Answer from: yesseniaroman21

The final angular velocity is w_f = 2.1994 rad/sec

The angular acceleration is $\alpha = 1.099 \ rad/sec^2$

Explanation:

From the question we are told that

The radius of each plate is $r = 30 \ cm = \frac{30}{100} = 0.3 \ m$

The mass of each plate is $m_p = 1 \ kg$

The angular speed of the spinning plate is $w = 0.7 \ rev \ per \ sec = 0.7 * 2 \pi = 4.3988 \ rad/sec$

From the law of conservation of momentum

L_i = L_f

Where L_i is the initial angular momentum of the system (The spinning and stationary plate ) which is mathematically represented as

L_i = I_1 w + 0

here I_1 is the moment of inertia of the spinning plate which mathematically represented as

$I_1 = \frac{m_pr^2}{2}$

and the zero signify that the stationary plate do not have an angular momentum as it is at rest at the initial state

L_f is the final angular momentum of the system (The spinning and stationary plate) , which is mathematically represented as

L_f = (I_1 + I_2 ) w_f

Where

I_2 is the moment of inertia of the second plate (This was stationary before but now it spinning due to the first pate ) and is equal to I_1

and w_f is the final angular speed

So we have

I_1 w = (I_1 + I_2)w_f

$\frac{m_p r^2}{2} * w = 2 * \frac{m_p r^2}{2} * w_f$

w = 2 * w_f

substituting values

4.3988 = 2 * w_f

$w_f = \frac{4.3988 }{2}$

w_f = 2.1994 rad/sec

The the rotational impulse-momentum theorem can be mathematially represented as

$\tau * \Delta t = 0.09891$

Where $\tau$ is the torque and $\Delta t$ is the change in time

So at $\Delta t = 2 \ sec$

$\tau = \frac{0.09891}{2}$

$\tau = 0.0995 \ Nm$

now the angular acceleation is mathematically represented as

$\alpha = 2 * \frac{\tau}{m_p * r^2 }$

substittuting values

$\alpha = 2 * \frac{0.0995}{1 * 0.3^2}$

$\alpha = 1.099 \ rad/sec^2$

Answer from: Quest

the earths atmosphere is made up of nitrogen and oxygen. solids can turn into gasses through a process called sublimation. sublimation is the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase.

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