Question

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What is the linear speed (in m/s) of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2

Answer 1

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Step-by-step explanation:

We are given that

Angular acceleration,
`\alpha=3.3 rad/s^2`

Diameter of the wheel, d=21 cm

Radius of wheel,
`r=(d)/(2)=(21)/(2)` cm

Radius of wheel,
`r=(21* 10^(-2))/(2) m`

1m=100 cm

Magnitude of total linear acceleration, a=
`1.7 m/s^2`

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,
`a_t=\alpha r`

`a_t=3.3* (21* 10^(-2))/(2)`

`a_t=34.65* 10^(-2)m/s^2`

Radial acceleration,
`a_r=(v^2)/(r)`

We know that

`a=√(a^2_t+a^2_r)`

Using the formula

`1.7=\sqrt{(34.65* 10^(-2))^2+((v^2)/(r))^2}`

Squaring on both sides

we get

`2.89=1200.6225* 10^(-4)+(v^4)/(r^2)`

`(v^4)/(r^2)=2.89-1200.6225* 10^(-4)`

`v^4=r^2* 2.7699`

`v^4=(10.5* 10^(-2))^2* 2.7699`

`v=((10.5* 10^(-2))^2* 2.7699)^{(1)/(4)}`

`v=0.418 m/s`

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s