A tire placed on a balancing machine in a service station starts from rest and t

Question

A tire placed on a balancing machine in a service station starts from rest and turns through 17.1 revolutions in 8.16 s before reaching its final angular speed. Calculate its angular acceleration.


Hint!!

This is just rotational kinematics, so we will use the rotational analog of our kinematic equations:

?f = ?i + ?i t + [ 1/2]at2

[instead of the linear version: xf = xi + vi t + [ 1/2]a t2]

We are given a rotational displacement ?? = ?f - ?i, and we are told that the tire starts at rest ( ?i = 0 ). So our equation simplifies to this:

?? = [ 1/2]at2

Which can be solved for alpha.

Finally there is the issue of units. We need our angular acceleration to be rad/s^2, and we are given our angular displacement in units of revolutions. Fortunately there are 2p rad in 1 revolution and we can easily convert.

Thank you for all your help!!

Explanation / Answer

angle covered on 17.1 revolutions= 2(17.1)= 107.388 radians apprx

Using kinematic equation

= it + 0.5t2 ( where = angular acceleration)

107.388 = 0 + 0.5(8.16)2 ( i = inital angular velcity = 0 rad/sec)

solving for =3.225 rad /s^2 apprx

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.