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yaroslaw [1]
4 years ago
8

A motorcycle, which has an initial linear speed of 8.0 m/s, decelerates to a speed of 2.2 m/s in 4.1 s. Each wheel has a radius

of 0.60 m and is rotating in a counterclockwise (positive) directions. What is
(a) the constant angular acceleration (in rad/s2) and
(b) the angular displacement (in rad) of each wheel?
Physics
1 answer:
SVETLANKA909090 [29]4 years ago
3 0

Answer:

Explanation:

Given

Initial linear speed v_1=8 m/s

initial angular velocity \omega _1=\frac{v_1}{r}=\frac{8}{0.6}=13.33 rad/s

Speed after 4.1 s is  v_2=2.2 m/s

\omega _2=\frac{2.2}{0.6}=3.66 rad/s

using \omega _2=\omega _1+\alpha t

where \alphais angular acceleration

3.66=13.33+\alpha \cdot 4.1

\alpha =-2.37 rad/s^2 i.e. clockwise

(b)angular displacement

\theta =\omega _1t+\frac{\alpha t^2}{2}

\theta =13.33\times 4.1-\frac{2.37\cdot 4.1^2}{2}

\theta =54.66-19.75

\theta =34.91 rad

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.

Let's calculate

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