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grandymaker [24]
4 years ago
13

A scooter has wheels with a diameter of 120 mm. What is the angular speed of the wheels when the scooter is moving forward at 6.

00 m/s?
Physics
1 answer:
nirvana33 [79]4 years ago
4 0

To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

v = r\omega \rightarrow \omega = \frac{v}{r}

Here,

v = Lineal velocity

\omega= Angular velocity

r = Radius

Our values are

v = 6/ms

r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m

Replacing to find the angular velocity we have,

\omega = \frac{6m/s}{0.06m}

\omega = 100rad/s

Convert the units to RPM we have that

\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})

\omega = 955.41rpm

Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm

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Answer:

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0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}

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v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}

As s(t) = vt + \frac{at^2}{2}. After t = t_1 + t_2, the car would have traveled a distance of

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Hence s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}

As t_1 = t_2 we can simplify s(t) = v_it_1

After t time, the train would have traveled a distance of s(t) = v_i(t_1 + t_2) = 2v_it_1

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Answer:

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a_c =  \frac{v^2}{r}
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For the car in this problem, a_c = 15.625 m/s^2 and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
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Read 2 more answers
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