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4 years ago

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A small cylinder rests on a circular turntable that is rotating clockwise at a constant speed. The cylinder is at a distance of

r = 12 cm from the center of the turntable. The coefficient of static friction between the bottom of the cylinder and the surface of the turntable is 0.45. What is the maximum speed vmax that the cylinder can have without slipping off the turntable?

1 answer:

r-ruslan [8.4K]4 years ago

6 0

Answer:

The maximum speed vmax = 0.727 m/s

Explanation

Given that

Distance ,r= 12 cm

Coefficient of static friction , μ= 0.45

Lets the mass of cylinder is m.

The friction force on the cylinder Fr = μ m g

Force due to rotation F

$F=mv^2/r$

v= Speed of rotation

The condition for no slipping

Fr= F

$\mu mg=mv^2/r$

$v=\sqrt{\mu rg}$

Now by pitting the values

$v=\sqrt{0.45\times 0.12\times 9.81}\ m/s$

v=0.727 m/s

So

The maximum speed vmax = 0.727 m/s

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Answer:

v=545.41 $\frac{m}{s}$

β=-25.93

Explanation:

Give the acceleration in 'x' and 'y' also the time can find the initial velocity using equation of uniform acceleration motion

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$a_{x}=5.10\frac{m}{s^{2}} \\v_{fx}=3780\frac{m}{s} \\v_{fx}=v_{ox}+a_{x}*t\\v_{ox}=v_{fx}-a_{x}*t\\v_{ox}=3780 \frac{m}{s}-5.1\frac{m}{s^{2} }*645s\\v_{ox}=490.5\frac{m}{s}$

For axis y

$a_{y}=7.30\frac{m}{s^{2}} \\v_{fy}=4470\frac{m}{s} \\v_{fy}=v_{oy}+a_{y}*t\\v_{oy}=v_{fy}-a_{y}*t\\v_{oy}=4470\frac{m}{s}-7.30\frac{m}{s^{2} }*645s\\v_{oy}=-238.5\frac{m}{s}$

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The direction is knowing when find the angle so

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Answer:

<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>

Explanation:

<u>Accelerated Motion</u>

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

$\displaystyle x=\frac{at^2}{2}$

Solving for a

$\displaystyle a=\frac{2x}{t^2}$

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Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

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Dividing by the weight of the box m.g

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