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andre [41]
2 years ago
5

Two objects have the same center point of the circle, but are located at different positions away from the center point. Each ob

ject is moving with uniform circular motion.
Which would describe the tangential speed of the objects?

both objects would have the same tangential speed
the object with the smaller radius has a faster tangential speed
the object with the larger radius has a faster tangential speed
both objects would have oscillating tangential speeds
Physics
1 answer:
Sergeeva-Olga [200]2 years ago
5 0

The object with the larger radius has a faster tangential speed. Tangential speed is related to both rotational speed and radial distance from the rotating axis.

<h3>What is uniform circular motion?</h3>

Uniform circular motion is a type of motion of a particle around a circle at a constant speed. The magnitude of the speed of the particle is constant.While the direction is changing continuously.

Tangential speed is related to both rotational speed and radial distance from the rotating axis.

The object with the larger radius has a faster tangential speed.

Hence, option C is correct.

To learn more about the uniform circular motion, refer to the link;

brainly.com/question/2285236

#SPJ1

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Answer:

a) 29062.125 N·m

b) 0 N·m

c) Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) T = 11186.02 N

Explanation:

We are given

Beam mass = 1975 kg

Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

Point of action of force = Midpoint for a uniform beam = length/2

∴ Point of action of force = 3/2 = 1.5 m

Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

∑M = 0 gives;

T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) Plugging in the values in the torque due to tension equation, we have;

3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125

Therefore, we make the tension force, T the subject of the formula hence

T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N

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