A tire 0.5m in radius rotates at a constant rate of 200 rev/min. find the speed and acceleration of a stone lodged in the tread of the tire (on its outer edge)
C = pi*2r = 3.14 * 1 = 3.14 m.
Vo = 200rev/min * 3.14m/rev. * 1min/60s = 10.5 m/s.
Since the velocity is constant, the acceleration is zero:
a = (V-Vo)/t = (10.5-10.5)/t = 0/t = 0.
To find the speed and acceleration of the stone lodged in the tread of the tire, we can use the formulas for linear speed and acceleration for circular motion.
First, let's find the linear speed of the stone. The linear speed of an object moving in a circular path is given by the formula:
v = r * ω,
where v is the linear speed, r is the radius of the circular path, and ω is the angular velocity.
In this case, the radius of the tire is given as 0.5m, and the angular velocity is given as 200 rev/min. We need to convert the angular velocity from revolutions per minute to radians per second. Since there are 2π radians in one revolution, we can convert the angular velocity as follows:
ω = (200 rev/min) * (2π rad/rev) * (1 min/60 s) = (200 * 2π) / 60 rad/s.
Substituting the values into the linear speed formula, we have:
v = 0.5m * [(200 * 2π) / 60] rad/s.
Simplifying the expression gives the linear speed, v.
Next, let's find the acceleration of the stone. The acceleration of an object moving in a circular path is given by the formula:
a = r * ω^2,
where a is the acceleration, r is the radius of the circular path, and ω is the angular velocity.
Substituting the values into the acceleration formula, we have:
a = 0.5m * [(200 * 2π) / 60]^2 rad/s^2.
Simplifying the expression gives the acceleration, a.
Now that we have derived the formulas and explained the steps to find the speed and acceleration, we can substitute the given values into the formulas to calculate the numerical values.