91影视

The following figure is given.

Apply the conservation energy equation at the initial position of the sled and when the sled is at an angle $蠒$. Consider the velocity of the sled when it is at an angle $蠒$is $v_1$. The initial velocity$v_0=0$

$$\begin{gathered} K_1+U_1=K_0+U_0 \\ \frac{1}{2}m{v}_1^2+mg{y}_1=\frac{1}{2}m{v}_0^2+mg{y}_0 \\ \cos \left(蠒\right)=\frac{y_1}{R} \\ {y}_1=R\cos \left(蠒\right) \\ {v}_0=0 \\ \frac{1}{2}m{v}_1^2+mgR\cos \left(蠒\right)=\frac{1}{2}m\left(0\right)+mgR \\ \frac{1}{2}m{v}_1^2+mgR\cos 蠒=mgR \\ \frac{1}{2}m{v}_1^2=mgR-mgR\cos 蠒 \\ {v}_1^2=2(gR-gR\cos 蠒) \\ {v}_1=\sqrt{2gR(1-\cos 蠒)} \end{gathered}$$

The following figure is given.

According to Newton's law, the net force on the sled is

$F_{net}=ma$

The acceleration of the sled while moving on the hill

$a=\frac{v^2}{R}$

The force acting on the sled is normal force (n) and force due to gravity along the axis. Therefore the net force, role="math" localid="1649389914026" $$\begin{gathered} F_{net}=mg\cos \left(蠒\right)-n \\ ma=mg\cos \left(蠒\right)-n \\ \text{Substitute(1)inequation(2)} \\ m\left(\frac{v^2}{R}\right)=mg\cos \left(蠒\right)-n \\ n=mg\cos \left(蠒\right)-\frac{m{v}^2}{R} \end{gathered}$$

The normal force (n) decreases as v increases. If $n=0$ the sled leaves the hill. But n can't be negative, so the fastest speed at which the sled stays on the hill is the speed The maximum speed$v_{max}$of the sled at which it stays on the hill is when $n鈮�0$.

Therefore the maximum speed of the sled is

localid="1649390700811" $v_{\max }=\sqrt{gR\cos 蠒}$

The following figure is given.

The speed of the sled at ${蠒}_{max}=\sqrt{2gR(1-\cos {蠒}_{max})}鈰�鈰�\left(1\right)$

The maximum speed of the sled at ${蠒}_{max}=\sqrt{gR\cos 蠒}鈰�鈰�\left(2\right)$

Equate the equation $\left(1\right)$and $\left(2\right)$

$$\begin{gathered} \sqrt{2gR(1-\cos {蠒}_{max})}=\sqrt{gR\cos {蠒}_{max}} \\ \text{Squaringonbothsides} \\ 2gR(1-\cos {蠒}_{max})=gR\cos {蠒}_{max} \\ 2\left(1-\cos {蠒}_{\max }\right)=\cos {蠒}_{\max } \\ 2-2\cos {蠒}_{\max }=\cos {蠒}_{\max } \\ 2=\cos {蠒}_{\max }+2\cos {蠒}_{\max } \\ 2=3\cos {蠒}_{\max } \\ \cos {蠒}_{\max }=\frac{2}{3} \\ {蠒}_{\max }={\cos }^{-1}\left(\frac{2}{3}\right) \\ {蠒}_{\max }=48掳 \end{gathered}$$