Question

A bicycle wheel of radius 0.3 m rolls down a hill without slipping. Its linear velocity increases constantly from 0 to 6 m/s in 2.7 s. What is its angular acceleration? Answer in units of rad/s 2 .Through what angle does the wheel turn in the 2.7 s? Answer in units of rad.How many revolutions does it make? Answer in units of rev.

Answer 1

`7.4 rad/s^(2)`, 27 rad, 4.3 rev

Step-by-step explanation:

The attachment contain the explanation.

Answer 2

(a)a= 2.222 m/s²

(b)angle= 26.973 rad

(c) 4.295 revolutions

Step-by-step explanation:

Given data

Radius=0.3 m

Linear velocity =0 m/s to 6 m/s

time taken=2.7 s

To find

(a) Angular acceleration in rad/s²

(b) Angle in radian

(c) Revolutions

Solution

For Part (a) of question

a) Use the equation acceleration = Alpha/R.........eq(i)

First find the linear acceleration

a = (v₂ – v₁)/(t₂ – t₁)

a= (6-0)/(2.7-0)

a= 2.222 m/s²

Put the value into the equation and solve for alpha:

angular acceleration=2.22/0.3

angular acceleration= 7.4 rad/s^2

For Part (b) of question

B) Use the equation өf = ө₀ + w₀t + (1/2)alpha×t²

This reduces to өf = (1/2)alpha×t²

Put the values into the equation

angle=(1/2)(7.4)(2.7)²

angle= 26.973 rad

For Part (c) of question

C) Convert the previous answer into revolutions

26.973 rad * (1rev/2π rad)

= 4.295 revolutions