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3 years ago

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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3 years ago

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Which of the following is the FINAL step in a forecasting system?

Lelechka [254]

Answer:

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3 years ago

A current I flows down a wire of radius a.

Helga [31]

Answer:

(a) $K = \frac{I}{2\pi a}$

(b) $J = \frac{I}{2\pi as}$

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

$K = \frac{dI}{dL}$

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

$dL = 2\pi r$

The radius of the wire = a

$dL = 2\pi a$

The surface current density $K = \frac{I}{2\pi a}$

(b) The current density is inversely proportional

$J \alpha s^{-1}$

$J = \frac{k}{s}$ ......(1)

k is the constant of proportionality

$I = \int\limits {J} \, dS$

$I = J \int\limits \, dS$ ........(2)

substituting (1) into (2)

$I = \frac{k}{s} \int\limits\, dS$

$I = k \int\limits^a_0 \frac{1}{s} {s} \, dS$

$I = 2\pi k\int\limits\, dS$

$I = 2\pi ka$

$k = \frac{I}{2\pi a}$

substitute $J = \frac{k}{s}$

$J = \frac{I}{2\pi as}$

7 0

3 years ago

b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force

Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

$= force \times radius$

$= f \times r$

And,

Torque is also

$= moment\ of\ inertia \times angular\ acceleration$

$= I \times \alpha$

So,

We can say that

$f \times r = I \times \alpha$

$f \times 0.05 = 0.2 \times 2$

0.05f = 0.4

f = 8 N

We simply applied the above formulas

8 0

3 years ago