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Elena-2011 [213]
3 years ago
8

Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for

problems like this one. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.130 m, an angular speed of 78.0 rad/s, and a moment of inertia of 1.25 kg·m2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of 6 during a time of 3.00 s. (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.
Physics
1 answer:
viva [34]3 years ago
5 0

Answer:

21.67 rad/s²

208.36538 N

Explanation:

\omega_f = Final angular velocity = \dfrac{1}{6}78=13\ rad/s

\omega_i = Initial angular velocity = 78 rad/s

\alpha = Angular acceleration

\theta = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875

Frictional force is given by

F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

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The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

<h3>What is mass?</h3>
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The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

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$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

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What heat transfer do sun heats earth go through
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991 food calories = 102073 × 4.2

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E = mgh

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