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Dovator [93]
3 years ago
9

A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b

icycle rests on. Assume that the rear tire is rotating at ω = 32k rad/s. What are the angular velocities of the two cylinders? Consider r1 = 460 mm and r2 = 46 mm.
Physics
1 answer:
mars1129 [50]3 years ago
8 0

Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

so

ω2  =  \frac{460\times 32}{34\times 2}

ω2  =  216.47

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A- 1000 m/s2<br> Xi-0m<br> Xf-0.75m<br> Vf-?
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Answer:

The final velocity of the object is,  v_{f} = 27 m/s    

Explanation:

Given,

The acceleration of the object, a = 1000 m/s²

The initial displacement of the object, x_{i} = 0 m

The final displacement of the object,  x_{f} = 0.75 m

The initial velocity of the object will be, v_{i} = o m/s

The final velocity of the object, v_{f} = ?

The average velocity of the object,

                                    v = ( x_{f} - x_{i} )/ t

                                      = 0.75 / t

The acceleration is given by the relation

                                     a = v / t

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Using the I equation of motion,

                                  v_{f} = u + at

Substituting the values

                                   v_{f} = 0 + 1000 x 0.027

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Hence, the final velocity of the object is,  v_{f} = 27 m/s          

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