Chapter 4, Problem 52P

To determine

The angle of inclination of the plane on which a skateboarder rolls down.

Answer to Problem 52P

Solution:

The skateboarder rolls down an incline at an angle 12⁰

Explanation of Solution

A skateboarder of mass m, rolls down an incline of length s, with an initial speed, $v_0$ in a time t. There is no friction between the incline and the skate board. The plane of the incline is at an angle $\theta$ to the horizontal.

The forces acting on the skateboarder is shown in the diagram below.

The weight mg of the skateboarder acts vertically downwards. The normal force $F_N$ acts perpendicular to the incline. The weight mg is resolved into two components, $mg\sin \theta \text{ and }mg\cos \theta$ parallel and perpendicular to the incline.

The skateboarder accelerates with an acceleration a downwards under the action of the component of weight $mg\sin \theta$.

    $F=mg\sin \theta$……(1)

By Newton’s second law,

    $F=ma$……(2)

From equations (1)and (2),

    $\begin{array}{c}ma=mg\sin \theta \\ a=g\sin \theta \end{array}$……(3)

If the skateboarder rolls down the incline of length s in a time t, use the equation of motion,

    $s=v_{0}t+\frac{1}{2}a{t}^2$

Using the expression for acceleration found in equation (3)in the above expression and simplifying for $\sin \theta$.

    $\begin{array}{c}s=v_{0}t+\frac{1}{2}a{t}^2\\ =v_{0}t+\frac{1}{2}g\sin \theta t^2\\ \sin \theta =\frac{2\left(s-v_{0}t\right)}{g{t}^2}\end{array}$

Given:

The initial speed of the skateboarder $v_0=2.0{\text{ m s}}^{-1}$

The length of the incline $s=18\text{ m}$

Time taken to slide down the incline $t=3.3\text{ s}$.

Formula used:

    $\sin \theta =\frac{2\left(s-v_{0}t\right)}{g{t}^2}$

Calculation:

Substitute the given values in the equation and calculate the value of $\sin \theta$.

    $\begin{array}{c}\sin \theta =\frac{2\left(s-v_{0}t\right)}{g{t}^2}\\ =\frac{2\left[\left(18\text{ m}\right)-\left(2.0\text{ m/s}\right)\left(3.3\text{ s}\right)\right]}{\left(9.8{\text{ m/s}}^2\right){\left(3.3\text{ s}\right)}^2}\\ =0.2136\end{array}$

Calculate the angle of incline $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(0.2136\right)\\ =12.3°\end{array}$