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3 years ago

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Find the approximate kinetic energy of a circular wheel of radius r and mass M that is rotating about its center at 2 cycles/s.

Assume the wheel’s mass is concentrated at the rim and the mass of the wheel’s spokes is negligible.

1 answer:

alukav5142 [94]3 years ago

6 0

Answer:

$8M(r\pi)^2$

Explanation:

First we convert 2 cycles/s to angular velocity knowing that each circle has an angle of 2π

$\omega = 2 * 2\pi = 4\pi rad/s$

Then we calculate the moment of inertia of the cylindrical shell, assuming there's no mass inside the wheel (only at the rim):

$I = mr^2 = Mr^2$

So the kinetic energy of this is

$E_k = I\omega^2/2 = Mr^2*(4\pi)^2/2 = 8M(r\pi)^2$

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Which one of the following types of training involves 10 to 30 minutes of high-intensity exercise?

Goshia [24]

Correct answer choice is:

C. Medium range

Explanation:

Medium range exercises are used to gain extra strength and fitness. Usually, heavyweights are used with less number of repetitions. These sort of exercises are mostly the hardest t do. All you need is to have a high level of motivation and stamina, which can be gained by running or cycling.

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A weight lifter applies an upward force of 1100 N while lowering a dumbbell

Paul [167]

Answer:

A

Explanation:

$work = force \times distance$

$work = 1100 \times 0.5$

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hope it helped a lot

pls mark brainliest with due respect .

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3 years ago

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

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3 years ago

Whats the net force

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Answer:

13n pushing left

Explanation:

1563 - 1550

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3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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