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fgiga [73]
3 years ago
10

A solid uniform cylinder is rolling without slipping. What fraction of its kinetic energy is rotational?

Physics
1 answer:
tester [92]3 years ago
5 0

Answer:

Explanation:

Let m be the mass of cylinder and r be the radius. It is moving with velocity v and angular velocity is ω. Let I be the moment of inertia of the cylinder.

I = 0.5 mr²

Total kinetic energy, T = 0.5 mv² + 0.5 Iω²

T = 0.5 (mv² + 0.5 mr²ω²)

v = rω

So, T = 0.5 (mv² + 0.5 mv²) = 0.75 mv²

Rotational kinetic energy is

R = 0.5 Iω² = 0.5 x 0.5 mr²ω²

R = 0.25 mv²

So, R / T = 0.25 / 0.75 = 1/3

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

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\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

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Replacing the previous equation with our values we have,

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\alpha = \frac{9.8436- 6.5973}{11}

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The tangential velocity then would be,

a = \alpha r

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Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

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3 years ago
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Answer:

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ahrayia [7]
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Answer:

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Explanation:

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