Question

A flywheel with a radius of 0.700 m starts from rest and accelerates with a constant angular acceleration of 0.500 rad/s2 .

A). Compute the magnitude of the resultant acceleration of a point on its rim at the start

B). Compute the magnitude of the resultant acceleration of a point on its rim after it has turned through 60.0 ∘.

C). Compute the magnitude of the resultant acceleration of a point on its rim after it has turned through 120.0

Answer #1

A)

at the start :

a_{t} = tangential acceleration = r = 0.7 x 0.5 = 0.35 m/s^{2}

a_{r} = radial acceleration = 0

so resultant acceleration = a = sqrt(a_{t}^{2} +
a_{r}^{2}) = sqrt(0.35^{2} + 0^{2})
= 0.35 m/s^{2}

b)

W_{i} = initial angular velocity = 0

W_{f} = final angular velocity = ?

= 0.5

= angular displacement = 60 degree = 1.05 rad

using the equation

W_{f}^{2} = W_{i}^{2} + 2

W_{f}^{2} = 02 + 2 (0.5) (1.05)

W_{f} = 1.025 rad/s

a_{r} = r W_{f}^{2} = (0.7)
(1.025)^{2} = 0.74 m/s^{2}

a_{t} = 0.35 m/s^{2}

so resultant acceleration = a = sqrt(a_{t}^{2} +
a_{r}^{2}) = sqrt(0.35^{2} +
0.74^{2}) = 0.82 m/s^{2}

c)

W_{i} = initial angular velocity = 0

W_{f} = final angular velocity = ?

= 0.5

= angular displacement = 120 degree = 2.1 rad

using the equation

W_{f}^{2} = W_{i}^{2} + 2

W_{f}^{2} = 02 + 2 (0.5) (2.1)

W_{f} = 1.45 rad/s

a_{r} = r W_{f}^{2} = (0.7)
(1.45)^{2} = 1.47 m/s^{2}

a_{t} = 0.35 m/s^{2}

so resultant acceleration = a = sqrt(a_{t}^{2} +
a_{r}^{2}) = sqrt(1.47^{2} +
0.35^{2}) = 1.51 m/s^{2}

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