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egoroff_w [7]
2 years ago
11

A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot

a 20cm long arrow parallel to the axle through the wheel at a speed of 6m/s. The arrow and the spokes are supposed to be thin. Calculate the maximum value of w ( in rad/second and in rev/second) so that the arrow just goes through without hitting any of the spokes. Does it matter where between the axle and the rim of the wheel you aim? If so, what is the best location?
Physics
1 answer:
spayn [35]2 years ago
3 0

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance <em>s</em><em> </em><em> </em>between the spokes along the rim is

s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}

The 20-cm arrow, traveling at 6 m/s, will travel its length in

t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}

The fastest speed that the wheel can spin without clipping the arrow is

v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}

The angular velocity \omega of the wheel is given by

\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}

In terms of rev/s, we can convert the answer above as follows:

23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

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An aluminum wing on a passenger jet is 35 m long when its temperature is 17°C. At what temperature would the wing be 3 cm (0.03
Mnenie [13.5K]

Answer:

53.32°C

Explanation:

Length of the aluminium wing = 35 m

Change in length of aluminium wing = 0.03 m

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We know that change in length is given by \Delta L=L\alpha \Delta T

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5 0
3 years ago
Determine the amount of work done on an ideal gas as it is heated in an enclosed thermally insulated cylinder topped with a free
sp2606 [1]

Answer:

W = 3/2 n (T₁- T₂)

Explanation:

Let's use the first law of thermodynamics

           ΔE = Q + W

in this case the cylinder is insulated, so there is no heat transfer

           ΔE = W

internal energy can be related to the change in temperature

            ΔE = 3/2 n K ΔT

we substitute

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as the work is on the gas it is negative

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3 years ago
Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
Fantom [35]

a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

I=\frac{1}{3}mL^2

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when \theta=10^{\circ} is

\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude mg

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

\tau = Fd

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

\tau =mg\frac{L}{2} cos \theta

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0
3 years ago
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