A cord is wrapped around the rim of a solid uniform wheel 0.280 m in radius and of mass 9.60 kg. a steady horizontal pull of 50.0 n to the right is exerted on the cord, pulling it off tangentially from the wheel. the wheel is mounted on frictionless bearings on a horizontal axle through its center
part a) compute the angular acceleration of the wheel
part b) compute the acceleration of the part of the cord that has already been pulled off the wheel
part c) find the magnitude of the force that the axle exerts on the wheel
part d) find the direction of the force that the axle exerts on the wheel
part e) which of the answers in parts a, b, c, and d would change if the pull were upward instead of horizontal?

The angular acceleration of the wheel and the translational acceleration of the cord depends on the mass and shape of the wheel and the applied.

Given;

Radius of the wheel, r = 0.280 m

Mass of the wheel, m = 9.60 kg

Horizontal pull force, F = 50.0 N

Direction of pull = To the right

Relating to the main wheel axle = Frictionless

Requires:

a. The angular acceleration of the wheel

Moment of inertia of the wheel, I = 1/2(m)(r)²

Therefore;

I = 1/2 X (9.60 kg) X (0.280)² = 0.37632 kgm²

The torque applied, τ = F·r

τ = I·α

Therefore;

F·r = I·α

Which gives;

α = (F X r)/I = (50.0N X 0.280m)/ 0.37632 kgm² = 37.202 rad/s²

The angular acceleration, α ≈ 37.202 rad/s²

b. To compute the acceleration of the part of the cord that has already been pulled off the wheel.

The section of the cord that was pulled out is moving to the right.

Translational acceleration, a = Angular acceleration, α × Radius, r

Therefore;

a ≈ 37.202 rad/s² × 0.280 m ≈ 10.416 m/s²

The acceleration of the part of the cord pulled out, a ≈ 10.416 m/s²

c. To find the magnitude of the force exerted by the axle on the wheel.

Solution:

There are two parts to the force the axle applies to the wheel;Horizontal component, Rₓ, and vertical component,

When the forces are balanced, we have;

∑Fₓ = The horizontal component, Rₓ + The pulling force applied, F = 0

Therefore;

Rₓ = -F = -50.0 N

∑fy = The weight of the wheel, W + The normal reaction, = 0

Therefore;

Ry= -W = -m·g ≈ -9.60 kg × (-1)× (9.81 m/s²) = 94.176 m/s²

The magnitude of the fore from the axle, R, is given as follows;

R = [tex]\sqrt{R^{2}_{x} + {R^{2}_{y}[/tex] = [tex]\sqrt{(-50.0 N})^{2} + \sqrt {(94.176N)^{2}[/tex] = 106.626 N

The magnitude of the force exerted by the axle, R ≈ 106.626 N

d. The direction of the force of the axle on the wheel.

Let θ Let be the direction of the fore in relation to the x-horizontal axis's axis.

we have;

tan θ = Ry/Rx = (94.176 m/s²)/(-50.0 N)

Which gives;

θ = arctan ((94.176 m/s²)/(-50.0 N)) = -61.879°

The direction of the force of the axle is, θ ≈-61.879° which is equivalent to 61.87° relative to the negative x-axis and 118.3° relative to the positive x-axis.

e. If the pull where upward instead of horizontal, the changes are;

Rx = 0,

Ry = W-F

tan θ = Ry/Rx

The magnitude and direction of the force, R will change'

Therefore;

The responses that will change are C and D.

Learn more about the Angular acceleration with the help of the given link:

https://brainly.com/question/15311727

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