Question

A tire 0.450 m in radius rotates at a constant rate of 190 rev/min. Find the speed of a small stone lodged in the tread of the tire (on its outer edge). (Hint: In one revolution, the stone travels a distance equal to the circumference of its path, 2π.) | m/s What is the acccleration of the stone? m/s2 O away from the center of the path tangent to the path toward the center of the path

Answer 1

Given:

Radius of tire R = 0.450 m

Angular speed ( N) (in revolution / min) = 190 rpm

( rpm = revolution per minute)

Step- 1

convert Angular speed in (rad/ s ) using relation

w ( rad/ s) = 2π N/60 ( In 1 revolution a point on tire will rotate 2 * π (radian) angle )

$\dpi{120} \bg_black \Rightarrow$ w= 2π * 190/60 rad/ s

$\dpi{120} \bg_black \Rightarrow$ w = 19.896 rad/ s

Step -2

Now use linear velocity ( m/s) and Angular velocity

( rad/s) relation:  

$\dpi{120} \bg_black \Rightarrow$ v = w* R

( As the stone is lodged ( fixed) on the outer periphery of tire. Hence the stone will also experience same Angular velocity as that of tire )

Speed of stone , v = wR

v = 19.896 * 0.450

v = 8.953 m/ s

Step - 3

Now use below relation to find centripetal accelaration of stone ( accelaration towards center )

Centripetal accelaration ,a = v^2/ R

a = 8.953^2/ 0.450

a = 178.12 m/ s^2

And this accelaration is directed towards the center of path.

( Note : As the tire is rotating at constant angular velocity , So the Tangential accelaration will be absent in it. )