Question

A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2. Compute the magnitude of the resultant acceleration (in m/s2) of a point on its rim after it has turned through 60.0°.

Answer 1

Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

$a_{t}$ = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

$a_{r}$ = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

$a = \sqrt{a_{t}^{2} + a_{r}^{2}}$

$a = \sqrt{0.18^{2} + 0.38^{2}}$

$a$ = 0.42 m/s²