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As a logging truck rounds a bend in the road, some logs come loose and begin to roll without slipping down the mountainside. The mountain slopes upward at 48.0 ∘ above the horizontal, and we can model the logs as solid uniform cylinders of mass 575 kg and diameter 1.20 m . Find the acceleration of the logs as they roll down the mountain. What is the friction force on the rolling logs? Is it kinetic or static?

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Answer

given,

mass of the uniform cylinder = 575 Kg

diameter = 1.20 m

angle of the slope = 48.0°

acceleration of the log = ?

[tex]I_{com}=\dfrac{1}{2}MR^2[/tex]

acceleration =

[tex]a =\dfrac{-g sin \theta}{1 + \dfrac{I_{com}}{MR^2}}[/tex]

[tex]a =\dfrac{-9.8\times sin 48^0}{1 + \dfrac{0.5MR^2}{MR^2}}[/tex]

[tex]a =\dfrac{-9.8\times sin 48^0}{1 + 0.5}[/tex]

a = -4.86 m/s²

negative direction of x- axis

frictional force

[tex]f_s = \dfrac{-I_{com} a}{R^2}[/tex]

[tex]f_s = \dfrac{-\dfrac{1}{2}MR^2 a}{R^2}[/tex]

[tex]f_s = -0.5 Ma[/tex]

[tex]f_s = -0.5\times 575 \times (-4.86) [/tex]

[tex]f_s = 1397.25 N[/tex]

c) It is a static friction

The acceleration of the logs down the hill will be at a constant rate given

that the hill has a constant slope.

Part 1: The acceleration of the logs as they roll down the mountain, a is approximately 4.86 m/s²

Part 2: The friction force on the rolling logs is approximately 1,397.3 N

Part 3: The friction force is static

Reasons:

Given parameters are;

Inclination of the slope, θ = 48°

Part 1;

The torque of the log, τ = M·g·sin(θ)·R

τ = I × α

[tex]I = \dfrac{2}{3} \cdot M \cdot R^2[/tex]

[tex]\alpha = \dfrac{M \cdot g \cdot sin(\theta) \cdot R}{ \dfrac{3}{2} \cdot M \cdot R^2} = \dfrac{2 \cdot g \cdot sin(\theta) }{ 3 \cdot R}[/tex]

a = α ×R

Therefore;

[tex]a = \dfrac{2 \cdot g \cdot sin(\theta) }{ 3 \cdot R} \times R = \dfrac{2 \cdot g \cdot sin(\theta) }{ 3 }[/tex]

Which gives;

[tex]The \ acceleration \ of \ the \ logs, \ a = \dfrac{2 \cdot g \cdot sin(48 ^{\circ}) }{ 3 } \approx 4.86 \, m/s^2[/tex]

The acceleration of the logs as they roll down the mountain, a ≈ 4.86 m/s²

Part 2:

Friction force, force, [tex]F_f[/tex] × R = [tex]I_{cm}[/tex] × α

[tex]I_{cm}[/tex] = The moment of inertia about the center of mass = [tex]\dfrac{1}{2} \cdot M \cdot R^2[/tex]

Therefore;

[tex]F_f = \dfrac{I_{cm} \cdot \alpha}{R} = \dfrac{\left(\dfrac{1}{2} \cdot M \cdot R^2 \right) \times \left(\dfrac{2 \cdot g \cdot sin(\theta) }{ 3 \cdot R}\right)}{R} = \dfrac{M \cdot g \cdot sin(\theta) }{ 3 }[/tex]

Therefore;

  • [tex]F_f = \dfrac{575 \times 9.81 \times sin(48^{\circ}) }{ 3 } \approx 1,397.3[/tex]

The friction force on the rolling logs, [tex]F_f[/tex] ≈ 1,397.3 N

Part 3

The logs are rolling without slipping, therefore, the friction force acting is static friction force

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