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bezimeni [28]
2 years ago
12

A disk-shaped platform has a known rotational inertia. The platform is mounted on a fixed axle and rotates in a horizontal plane

, as shown above. A student wishes to determine the frictional torque exerted on the platform by the axle as the platform rotates. The student has access to equipment that would usually be found in a school physics laborator
Physics
1 answer:
Nataly [62]2 years ago
5 0

The frictional torque exerted on the platform by the axle as the platform rotates will be;

\rm T_f \theta =\frac{1}{2} I [\omega^2-\omega_0^2]

<h3>What is torque?</h3>

Torque is the force's twisting action about the axis of rotation. Torque is the term used to describe the instant of force. It is the rotational equivalent of force. Torque is a force that acts in a turn or twist.

The amount of torque is equal to force multiplied by the perpendicular distance between the point of application of force and the axis of rotation.

Work done by the frictional torque = Change in the rotational kinetic energy of the wheel

\rm T_f \theta =\frac{1}{2} I [\omega^2-\omega_0^2]

Where,

\rm T_f is the frictional torque

\rm \omega is the final angular velocity

\rm  \omega_0 is the initial angular velocity

\rm \theta is the angular displacement

Hence, the frictional torque exerted on the platform by the axle as the platform rotates will be;

\rm T_f \theta =\frac{1}{2} I [\omega^2-\omega_0^2]

To learn more about the torque, refer to the link;

brainly.com/question/6855614

#SPJ1

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Answers:

a) 0.80 kg

b) 2.12 s

c) 1.093 m/s^{2}

Explanation:

We have the following data:

k=7 N/m is the spring constant

A=12.5 cm \frac{1 m}{100 cm}=0.125 m is the amplitude of oscillation

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Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

U_{o}+K_{o}=U_{f}+K_{f} (1)

Where:

U_{o}=k\frac{A^{2}}{2} is the initial potential energy

K_{o}=0  is the initial kinetic energy

U_{f}=k\frac{x^{2}}{2} is the final potential energy

K_{f}=\frac{1}{2} m V^{2} is the final kinetic energy

Then:

k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2} (2)

Isolating m:

m=\frac{k(A^{2}-x^{2})}{V^{2}} (3)

m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}} (4)

m=0.80 kg (5)

<h3>b) Period</h3>

The period T is given by:

T=2 \pi \sqrt{\frac{m}{k}} (6)

Substituting (5) in (6):

T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}} (7)

T=2.12 s (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration a_{max} is when the force is maximum F_{max}, as well :

F_{max}=m.a_{max}=k.x_{max} (9)

Being x_{max}=A

Hence:

m.a_{max}=kA (10)

Finding a_{max}:

a_{max}=\frac{kA}{m} (11)

a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg} (12)

Finally:

a_{max}=1.093 m/s^{2}

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3 years ago
Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an angl
alexandr1967 [171]

Answer:

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Explanation:

From law of reflection i=r.

So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.  

Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.

If you consider triangle AOB, one angle is ∠AOB=90°

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From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°

 

So, the second incident angle will be 54.6°

Hence, the second reflected angle will be 54.6 degrees.

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Answer:

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2 years ago
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