A wheel of 40cm radius rotates on a stationary axle. It isuniformly speeded up from rest to a speed of 900 rpm in a time of20s. Find (a) the constant angular acceleration of the wheel and(b) the tangential acceleration of a point of a point on itsrim.
(a) convert 900 rpm to radians per second and divide by 20s. That is usually called alpha, in radians/s^2
(b) tangential acceleration = R*(alpha)
This is basic stuff. Show your work if additional assistance is needed
To find the constant angular acceleration of the wheel, we can use the formula:
Angular acceleration (α) = (Final angular velocity - Initial angular velocity) / Time
Given:
Initial angular velocity (ω₁) = 0 (as the wheel starts from rest)
Final angular velocity (ω₂) = 900 rpm = (900 * 2π) / 60 rad/s (1 revolution = 2π radians)
Time (t) = 20 s
Substituting the values into the formula:
α = ((900 * 2π) / 60 - 0) / 20
Simplifying,
α = (900 * 2π) / 60 * 20
The answer gives the constant angular acceleration of the wheel.
To find the tangential acceleration of a point on the rim of the wheel, we can use the formula:
Tangential acceleration (at) = Angular acceleration (α) * Radius of the wheel
Given:
Radius of the wheel (r) = 40 cm = 0.4 m
Substituting the values into the formula:
at = α * r
Simplifying,
at = ((900 * 2π) / 60 * 20) * 0.4
The answer gives the tangential acceleration of a point on the wheel's rim.
Please note that the SI unit for angular acceleration and tangential acceleration is radians per second squared (rad/s²).