The equation for the circumference of a circle is c = 2πr, where r is the radius of the circle. If the circle was a tire it would roll through a distance of 2πr for every revolution.
The tangential velocity is n2πr where n is the number of revolutions per second. That’s the distance in space the rotating spot in question is carried through by the rotation in one second.

Question 9. A tire of radius .500 m turns through 5.00 revolutions in one second. If this motion is steady, what is the tangential velocity of a spot on the rim of the tire?

Question

The equation for the circumference of a circle is c = 2πr, where r is the radius of the circle. If the circle was a tire it would roll through a distance of 2πr for every revolution.
The tangential velocity is n2πr where n is the number of revolutions per second. That’s the distance in space the rotating spot in question is carried through by the rotation in one second.

Question 9. A tire of radius .500 m turns through 5.00 revolutions in one second. If this motion is steady, what is the tangential velocity of a spot on the rim of the tire?

shamim · Answer

V=2*pi*r*n=2*pi*0.5*5=?