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3 years ago

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After fixing a flat tire on a bicycle you give the wheel a spin. If its initial angular speed was 6.36 rad/s and it rotated 14.7

revolutions before coming to rest, what was its average angular acceleration (assuming that the angular acceleration is constant)

1 answer:

lesya692 [45]3 years ago

4 0

To solve this problem we will apply the concepts related to the cinematic equations of angular motion. On these equations, angular acceleration is defined as the squared difference of angular velocity over twice the radial displacement. This is mathematically:

$\alpha = \frac{\omega^2-\omega_0^2}{2\theta}$

Our values are,

$\text{Initial angular velocity} = \omega_0 =6.36 rad/s$

$\text{Final angular velocity} = \omega =0$

$\text{Angular displacement} = \theta = 14.7rev = 29.4\pi rad$

Replacing,

$\alpha = \frac{- 6.36^2}{29.4\pi}$

$\alpha = -0.43rad/s^2$

Therefore the angular acceleration is $-0.43rad/s^2$

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Read 2 more answers

A skateboarder is trying to skate horizontally off a building and jump over a pool as shown! The pool is 25.0m wide and the roof

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Answer:

a) t = 2.55s

b) $v_{0x} = 9.80 m/s$

c) yes

Explanation:

In order to solve this problem, we can start by drawing a sketch of the situation so we can better visualize what the problem is about (see attached picture).

a)

For the first question. We are talking about a movement in two dimensions. So on the first question they are asking us for vertical movement. It will be uniformly accelerated, so we can use the following formula:

$y_{f}=y_{0}+V_{0y}t+\frac{1}{2}at^{2}$

We know the following:

$y_{f}=0$

$y_{0} = 32m$

$V_{0y}=0$

t=?

$a=-9.81 m/s ^{2}$

With this data, we can simplify our equation, so we end up with:

$y_{0}+\frac{1}{2}at^{2}=0$

so we can now substitute the data we know and solve for t:

$32m-\frac{1}{2}(9.81 m/s^{2})t^{2}=0$

$(-4.905 m/s^{2})t^{2}=-32m$

$t^{2} = \frac{-32m}{-4.905 m/s^{2}}$

$t=\sqrt{6.52s^{2}}$

t = 2.55 s

b)

For part b, since we are talking about horizontal movement and we are neglecting drag, this means that his horizontal speed will be constant. So we can use the following formula:

$V_{x}=\frac{x}{t}$

we know he most move a horizontal distance of 25 meters in a time of 2.55s so we get:

$V_{x} = \frac{25m}{2.55s}$

$V_{x}=9.80 m/s$

c) for part c, we can do the conversion between miles per hour to meters per second like this:

$\frac{27mi}{1hr}*\frac{1hr}{3600s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}$

so the given initial speed is equivalent to:

12.07 m/s

this is greater than the minimum 9.80 m/s we need, so the skater will clear the pool at this speed.

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