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frosja888 [35]
3 years ago
5

A string is wrapped around a uniform cylinder of mass M and radius R. The cylinder is released from rest with the string vertica

l and its top end tied to a fixed bar. A uniform disk, cylindrical in shape, of mass M and radius R is suspended from a string. The top end of the string is attached to a fixed bar. The string extends down through a distance h before reaching the disk, whereupon the string winds around the cylindrical body of the disk. (a) Show that the tension in the string is one-third the weight of the cylinder. (Submit a file with a maximum size of 1 MB.) (b) Show that the magnitude of the acceleration of the center of gravity is 2g/3. (Submit a file with a maximum size of 1 MB.) (c) Show that the speed of the center of gravity is 4gh 3 1/2 after the cylinder has descended through distance h. Verify your answer to (c) with the energy approach. (Submit a file with a maximum size of 1 MB.)
Physics
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

a.) W/3,  b.)2g/3   c.) (4gh/3)^0.5

Explanation:

First we have to find tension in terms of torque. To do that we have to find the moment of inertia of a rigid cylinder. From Wikipedia I get:

I = 0.5mr^2

We also know the equation for torque. Let T be the tension, r be the distance, I is the moment of inertia and alpha is angular acceleration (ignore theta because it is perpendicular)

torque = I*\alpha = T*r

We can then substitute a/r for α

Therefore we get:

I*\frac{a}{r}=T*r

Isolating T and substitute the moment of inertia in for I we get

T = [0.5mr^2]a/r^2= 0.5ma

There are two known forces acting on the cylinder, gravity and tension. The sum of these two forces gives us mass times acceleration (Newton's second law)

F = ma => mg - 0.5ma = ma\\g=\frac{3}{2}a\\a = \frac{2}{3}g

This allows us to plug acceleration back into Newton's Second Law:

mg - T = m[\frac{2}{3}g]\\ T = \frac{1}{3} mg = \frac{1}{3}w

w = the weight

For part b, we solved in part a:

a = \frac{2}{3}g

For part c, we use the conservation of energy. We know that the sum of energy in the system is zero.

mgh + \frac{1}{2} mv^2 + \frac{1}{2}I(omega)^2 = 0\\(omega) = (\frac{v}{r})^2\\ I = \frac{1}{2} mr^2\\-gh +\frac{1}{2} v^2+\frac{1}{4} v^2 = 0\\gh = \frac{3}{4} v^2\\v = \sqrt{ \frac{4}{3}gh }

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What's is the kinetic energy of a .235-kg baseball thrown at 50.0m/s
san4es73 [151]

Answer:

34.51

Explanation:

k=1/2mv² is the kenetic energy equation to fill is in

k=[1/2(0.235)×50]²

6 0
3 years ago
A boy throws a ball with an initial velocity of 19.6 m/s. What maximum height does the ball reach?
Wewaii [24]
<h2>Hello!</h2>

The answer is: 19.59 m

<h2>Why?</h2>

Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.

When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:

We will use the following formula:

Vf^2=Vi^2+2*g*s

Where:

Vf= Final velocity = 0

Vi= Initial velocity = \frac{19.6m}{s}

g = Gravity Acceleration = \frac{9.81m}{s^{2} }

s = Traveled distance

0=19.6^2+2*-9.81*s\\s=\frac{19.6^2}{2*9.81}=\frac{384.16}{19.62}=19.59m

Have a nice day!

8 0
3 years ago
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
3 years ago
A hamster in it's ball starts at rest and accelerates to 3ms1 in 6 seconds.
taurus [48]

Answer:9m

Explanation:

Ball starts from rest . Time taken = 6 seconds. Distance travelled by ball. ∴Distance travelled = 9 m

Hope it helps you

Good luck

7 0
3 years ago
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Vlad [161]

Answer:

1 different

Explanation:

2 light

3 sunny

4 a shadow

5dark

<h2 />
8 0
3 years ago
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