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Sagot :
Solution :
Given :
Mass, m = 7.80 kg
Radius, r = 0.280 m
Force, F = 34 N
a). St = I x a
[tex]$34 \times r = 7.8 r^2 \times \frac{a}{2}$[/tex]
[tex]$a=\frac{68}{7.8 \times 0.280}$[/tex]
[tex]$=31.13 \ rad/s^2$[/tex]
b). The acceleration of the part of the chord is
= 31.13 x 0.280
[tex]$=8.7164 \ m/s^2$[/tex]
c). Magnitude of the force is = [tex]$\sqrt{34^2 + (7.8 \times 9.81)^2}$[/tex]
= 83.73 N
d). The direction of force is = arctan (7.8 x 9.81)/34
= 2.25
e). The answers in part (c) and (d) will change when the pull of force is given in the upward direction instead of giving in the horizontal direction
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.
The correct responses are;
- a. 31.14 rad/s²
- b. 8.72 m/s²
- c. 83.73 N
- d. 113.96°
- e. C and D
Reasons:
Radius of the wheel, r = 0.280 m
Mass of the wheel, m = 7.80 kg
Horizontal pull force, F = 34.0 N
Direction of pull = To the right
Bearing on the central wheel axle = Frictionless
Requires:
a. The angular acceleration of the wheel
Solution:
Moment of inertia of the wheel, [tex]\displaystyle I = \mathbf{\frac{1}{2} \cdot m \cdot r^2}[/tex]
Therefore;
[tex]\displaystyle I = \frac{1}{2} \times 7.80 \, kg. \times (0.280 \, m)^2 = \mathbf{0.30576 \, kg \cdot m^2}[/tex]
The torque applied, τ = F·r
τ = I·α
Therefore;
F·r = I·α
Which gives;
[tex]\displaystyle \alpha = \mathbf{ \frac{F \times r}{I}} = \frac{34.0 \, N \times 0.280 \, m}{0.30576 \, kg\cdot m^2} \approx 31.14 \, rad/s^2[/tex]
- The angular acceleration, α ≈ 31.14 rad/s²
b. To compute the acceleration of the part of the cord that has already been pulled off the wheel.
Solution:
The part of the cord pulled out is translating to the right
Translational acceleration, a = Angular acceleration, α × Radius, r
Therefore;
a ≈ 31.14 rad/s² × 0.280 m ≈ 8.72 m/s²
- The acceleration of the part of the cord pulled out, a ≈ 8.72 m/s²
c. To find the magnitude of the force exerted by the axle on the wheel.
Solution:
The force exerted by the axle on the wheel has two components; Horizontal component, Rₓ, and vertical component, [tex]\mathbf{R_y}[/tex]
When the forces are balanced, we have;
∑Fₓ = The horizontal component, Rₓ + The pulling force applied, F = 0
Therefore;
Rₓ = -F = -34.0 N
[tex]\sum F_y[/tex] = The weight of the wheel, W + The normal reaction, [tex]R_y[/tex] = 0
Therefore;
[tex]R_y[/tex] = -W = -m·g ≈ -7.80 kg × (-1)× (9.81 m/s²) = 76.518 m/s²
The magnitude of the fore from the axle, R, is given as follows;
[tex]R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-34.0 \, N)^2 + (76.518 \, N)^2} \approx \mathbf{83.73 \, N}[/tex]
- The magnitude of the force exerted by the axle, R ≈ 83.73 N
d. The direction of the force of the axle on the wheel.
Solution:
Let θ represent the direction of the fore relative to the horizontal x-axis, we have;
[tex]\displaystyle tan(\theta) = \frac{R_y}{R_x} = \mathbf{\frac{76.518}{-34}}[/tex]
Which gives;
[tex]\displaystyle \theta = arctan\left (\frac{76.518}{-34} \right) \approx -66.04^{\circ}[/tex]
- The direction of the force of the axle is, θ ≈ -66.04° which is equivalent to 66.04° relative to the negative x-axis and 113.96° relative to the positive x-axis.
e. If the pull where upward instead of horizontal, the changes are;
Rₓ = 0,
[tex]R_y[/tex] = W - F
[tex]\displaystyle tan(\theta) = \mathbf{\frac{R_y}{R_x}}[/tex]
The magnitude and direction of the force, R will change'
Therefore;
- The responses that will change are C and D
Learn more about moment of inertia here:
https://brainly.com/question/6545003
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