Question

A bicycle with 0.80-m-diameter tires is coasting on a level road at 5.6 m/s . A small blue dot has been painted on the tread of the rear tire.

a) What is the angular speed of the tires?
b) What is the speed of the blue dot when it is 0.80m above the road?
c) What is the speed of the blue dot when it is 0.40m above the road?

Answer 1

a)
`\omega=14\ rad.s^(-1)`

b)
`v=11.2\ m.s^(-1)`

c)
`v_r=5.6√(2)\ m.s^(-1)`

Step-by-step explanation:

Given:

• diameter of bicycle-tyre,
  `d=0.8\ m`

• linear speed of tyre,
  `v_t=5.6\ m.s^(-1)`

a)

angular speed of the tyre:

We know the relation between angular and linear speed as:

`\omega=(v)/(r)`

here
`r=` radius of the wheel

`\omega=(5.6)/(0.4)`

`\omega=14\ rad.s^(-1)`

b)

Speed of blue dot on the periphery of the tyre when the dot is 0.8 meter above the ground:

`v=v_t+r.\omega`

this velocity will be the sum of the linear velocity of the tyre and the tangential velocity of the point.

`v=5.6+5.6`

`v=11.2\ m.s^(-1)`

c)

When the dot is at a height of 0.4m it is either going down the rotation or it is going up the rotation which is normal to the linear velocity of the tyre, hence resultant effect is:

`v_r=√(5.6^2+5.6^2)`

`v_r=5.6√(2)\ m.s^(-1)`

Answer 2

14 rad/s

11.2 m/s

7.91959 m/s

Step-by-step explanation:

d = Diameter = 0.8 m

r = Radius

v = Velocity = 5.6 m/s

`\omega` = Angular speed

Velocity is given by

`v=r\omega\\\Rightarrow \omega=(v)/(r)\\\Rightarrow \omega=(5.6)/((0.8)/(2))\\\Rightarrow \omega=14\ rad/s`

The angular speed is 14 rad/s

Above the ground the speed of the dot would be

`v_a=v+r\omega\\\Rightarrow v_a=5.6+5.6\\\Rightarrow v_a=11.2\ m/s`

The speed of the dot is 11.2 m/s

The resultant of the velocities will be

`R=√(5.6^2+5.6^2)\\\Rightarrow R=7.91959\ m/s`

The velocity of the dot is 7.91959 m/s