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MrRa [10]
3 years ago
11

A uniform plank is 3.0 m long and has a mass of 10 kg. It is secured at its left end in a horizontal position to be used as a di

ving platform. To keep the plank in equilibrium, the point of support must supply: a. an upward force and a clockwise torque b. a downward force and a clockwise torque c. an upward force and a counter-clockwise torque d. a downward force and a counter-clockwise torque e. none of these

Physics
1 answer:
garik1379 [7]3 years ago
7 0

To develop the problem it is necessary to take into account the concepts related to Torque and sum of moments.

By torque it is understood that

\tau = F*d

Where,

F= Force

d = Distance

The value of the given Torque acts from the center of mass causing it to rotate clockwise.

The Force must then be located at the other end down to make a movement opposite the Torque in the center of mass.

I enclose a graph that allows us to understand the problem in a more didactic way.

The correct answer is D.

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To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

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\omega_f = \omega_0 + \alpha t

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\omega_0 =Initial Angular velocity

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The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

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Explanation:

Answer

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An object moves 4 meters north, then 3 meters east. What is the displacement of the object?
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The displacement of the object that moved 4 meters north, then 3 meters east is 5 meters.

<h3>What is displacement?</h3>

Displacement is change in the position of an object. It can be described as the shortest distance between the initial and final position of an object.

The displacement of the object is calculated as follows;

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Learn more about displacement here: brainly.com/question/2109763

#SPJ2

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2 years ago
Read 2 more answers
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