Question

A hill with a slope of 35° is situated at the top of a cliff that is 44݉ high. A solid disk, beginning at rest, rolls down the hill. The length of the incline is 12.55݉. If a kid is standing 20݉ away from the cliff, will the disk hit him?&

Answer 1

A hill with a slope of $35{}^{\circ}$ is situated at the top of a cliff that is 44m high. A solid disk, beginning at rest, rolls down the hill. The length of the incline is 12.55m. If a kid is standing 20m away from the cliff, will the disk hit him?

Solution:

$\alpha = 35{}^{\circ}$ – slope of the hill;

$H = 44m$ – high of the cliff;

$L = 12.55m$ – length of the incline;

$s = 20m$ – distance from the kid to the cliff;

$R$ – radius of the disk;

$S$ – horizontal distance that the disk traveled

Law of conservation of energy during disk is rolling down the hill:

$W_{potential} = W_{kinetic}$

$$mgL \sin \alpha = \frac{mV^2}{2} + \frac{J\omega^2}{2}$$

$J$ – moment of inertial of the disk, $J = \frac{mR^2}{2}$

$\omega$ – angular velocity of the disk, $\omega = \frac{V}{R}$

$$\frac{mV^2}{2} + \frac{J\omega^2}{2} = \frac{mV^2}{2} + \frac{mR^2}{4} \cdot \left(\frac{V}{R}\right)^2 = \frac{mV^2}{2} + \frac{mV^2}{4} = \frac{3mV^2}{4}$$

(2)in(1):

$$mgL \sin \alpha = \frac{3mV^2}{4}$$ $$4gL \sin \alpha = 3V^2$$ $$V = 2\sqrt{\frac{gL \sin \alpha}{3}} = 2\sqrt{\frac{9.8\frac{m}{S^2} \cdot 12.55 \cdot \sin 35{}^{\circ}}{3}} = 9.7\frac{m}{s}$$

Equations of motion for the disk after the detachment along the X-axis and Y-axis $(\beta = 90{}^{\circ} - \alpha)$.

$$x: S = Vt \cos \beta = Vt \cos(90{}^{\circ} - \alpha) = Vt \sin \alpha$$ $$t = \frac{S}{V \sin \alpha}$$ $$y: H = Vt \sin \beta + \frac{gt^2}{2} = Vt \cos \alpha + \frac{gt^2}{2}$$

(4)in(5):

$$H = V \cos \alpha \cdot \frac{S}{V \sin \alpha} + \frac{g}{2} \cdot \left(\frac{S}{V \sin \alpha}\right)^2$$ $$2HV^2 \sin^2 \alpha = \frac{2S \cdot V^2 \sin^2 \alpha}{\tan \alpha} + gS^2$$ $$2HV^2 \sin^2 \alpha = 2S \cdot V^2 \sin \alpha \cdot \cos \alpha + gS^2$$

We have quadratic equation:

$9.8 \cdot S^{2} + 88.4 \cdot S - 2720 = 0$

$S = 12.75m$

12.75m $\neq 20m \Rightarrow$ the disk will not hit the kid

Answer: the disk will not hit the kid (horizontal distance that traveled the disk is equal $12.75\mathrm{m}$ )