A wheel with a radius of 1.5m rotates at uniform angular speed. If a point on the rim of the wheel has a centripetal acceleration of 1.2 m/s/s whats the points tangential speed?
a= v²/R
v=sqrt(aR)
To find the point's tangential speed, we need to use the formula for centripetal acceleration and the relationship between tangential speed and angular speed.
The formula for centripetal acceleration is given by:
a = ω^2 * r
where:
a is centripetal acceleration,
ω (omega) is angular speed, and
r is the radius of the wheel.
In this case, the centripetal acceleration (a) is given as 1.2 m/s², and the radius (r) is 1.5 m.
Rearranging the formula, we get:
ω^2 = a / r
Now, we need to find ω. To do that, we take the square root of both sides:
ω = √(a / r)
Substituting the given values, we have:
ω = √(1.2 / 1.5) ≈ 0.8717 rad/s
The relationship between tangential speed (v) and angular speed (ω) is given by:
v = ω * r
In this case, we already know ω (0.8717 rad/s) and r (1.5 m), so we can calculate the tangential speed (v):
v = 0.8717 rad/s * 1.5 m ≈ 1.3075 m/s
Therefore, the point on the rim of the wheel has a tangential speed of approximately 1.3075 m/s.