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# Tutor profile: Anibal S.

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Anibal S.
AP Physics Teacher
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A disc of mass $$M$$ and radius $$R$$ rotates about a fixed axis with angular velocity $$\omega$$. Give expressions for the rotational kinetic energy of the disc if it rotates about an axis a) through its geometric center b) through a point halfway between the center and the rim of the disc.

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Anibal S.

In either case the rotational kinetic energy is calculated with the formula $$K_R = \frac 1 2 I \omega^2$$. The only difference is the expression for the moment of inertia $$I$$. a) In the first case $$I = \frac 1 2 M R^2$$, therefore $$K_R = \frac 1 2 (\frac 1 2 M R^2) \omega^2 = \frac 1 4 M R^2 \omega^2$$. b) In the second case, according to the parallel axes theorem, $$I = I_{CM} + M D^2$$, where in this case $$D = \frac 1 2 R$$. So $$I = \frac 1 2 M R^2 + M(\frac 1 4 R^2) = \frac 3 4 M R^2$$. Then the rotational kinetic energy will be $$K_R = \frac 1 2 (\frac 3 4 M R^2) \omega^2 = \frac 3 8 M R^2 \omega^2$$

### Subject:Calculus

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Question:

Derive the function $$f(x) = x^3 \sin {2x}$$.

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Anibal S.

According to the product rule, $$f'(x) = \frac d {dx} x^3 \sin {2x} + x^3 \frac d {dx} \sin {2x}$$ $$=3x^2 \sin {2x} + x^3 2 \cos {2x}$$ $$= 3x^2 \sin {2x} + 2x^3 \cos {2x}$$ Notice that in step two we make use of the chain rule on the sine function.

### Subject:Physics

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Question:

You throw a projectile horizontally with a velocity $$v$$ from a height $$h$$ above the ground. The projectile lands a horizontal distance $$d$$ from the point you threw it. If you now throw the projectile from a height $$2h$$, how is the horizontal displacement affected by this change?

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Anibal S.

The new horizontal displacement $$d_2$$ depends on the time of flight $$t_2$$ through the relationship $$d_2 = v t_2$$. The time, however, is depends on the vertical displacement through the relationship $$2h = \frac 1 2 g t_2^2$$. Therefore, $$t_2 = \sqrt { \frac {2(2h)} g }$$. If we consider that the time for the first launch is $$t = \sqrt { \frac {2h} g }$$, it is clear that $$t_2 = \sqrt 2 t$$ and therefore $$d_2 = \sqrt 2 v t$$. In other words, the horizontal displacement increases by a factor of $$\sqrt 2$$.

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