Question

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?

Answer 1

Answer:

Acceleration, a = 4.85$m/s^{2}$

Friction Force, F = 1395.8 N

Solution:

As per the question:

Angle between the road and the horizontal, $\theta = 48.0^{\circ}$

Mass of the cylinder, m = 575 kg

Diameter of the cylinder, d = 1.20 m

Radius of the cylinder, $R = \frac{d}{2} = \frac{1.20}{2} = 0.60\ m$

Now,

To calculate the acceleration of the log:

Firstly, using parallel axis theorem for the moment of inertia:

Moment of inertia about the log's point of contact having slope, $I = \frac{3}{2}mR^{2}$

Also,

About the point of contact, torque is given by:

$\tau = mgRsin\theta$

Also, we know that:

$\tau = I\alpha$

where

$\alpha$ = angular acceleration

Therefore,

$I\alpha = mgRsin\theta$

$\alpha = \frac{mgRsin\theta}{I}$

$\alpha = \frac{mgRsin\theta}{\frac{3}{2}mR^{2}}$

$\alpha = \frac{2gsin\theta}{3R}$

$\alpha = \frac{2\times 9.8\times sin48^{\circ}}{3\times 0.6} = 8.09\ rad/s^{2}$

Linear acceleration:

$a = \alpha R = 8.09\times 0.6 = 4.85\ m/s^{2}$

Now,

To calculate the friction force:

Here, the friction force gives the torque that provides the angular acceleration about the center:

$F\times R = \alpha I_{CM}$

where

F = friction force

$I_{CM} = \frac{1}{2}mR^{2}$ = Moment of Inertia about the center

$F = \frac{\alpha I_{CM}}{R}$

$F = \frac{\frac{2gsin\theta}{3R}\times \frac{1}{2}mR^{2}}{R}$

$F = \frac{mgsin\theta}{3} = [tex]F = \frac{575\times 9.8\times sin48.0^{\circ}}{3} = 1395.8\ N$

Answer 2

Answer

given,

mass of the uniform cylinder = 575 Kg

diameter = 1.20 m

angle of the slope = 48.0°

acceleration of the log = ?

$I_{com}=\dfrac{1}{2}MR^2$

acceleration =

$a =\dfrac{-g sin \theta}{1 + \dfrac{I_{com}}{MR^2}}$

$a =\dfrac{-9.8\times sin 48^0}{1 + \dfrac{0.5MR^2}{MR^2}}$

$a =\dfrac{-9.8\times sin 48^0}{1 + 0.5}$

a = -4.86 m/s²

negative direction of x- axis

frictional force

$f_s = \dfrac{-I_{com} a}{R^2}$

$f_s = \dfrac{-\dfrac{1}{2}MR^2 a}{R^2}$

$f_s = -0.5 Ma$

$f_s = -0.5\times 575 \times (-4.86)$

$f_s = 1397.25 N$

c) It is a static friction