Question

Problem 3 A block of mass m slides down a frictionless incline. The block is released a height h above the bottom of the loop. The bottom of the loop is circular with radius R. a) What is the force of the track on the block at point A? Express your answer in terms of m, g, h, and R. b) What is the force of the track on the block at point B? Express your answer in terms of m, g, h, and R. c) At what speed does the block leave the track? Express your answer in terms of h and R. d) How far away from point A does the block land on level ground? You might find it simpler expressing your answer in terms of v, g, and h'. e) Sketch the potential energy of the block. Indicate the total energy on the sketch. 145 B h' † A →X

Answer 1

Refer to the diagrams:

h R 1450 9450 In'= R(1-6545) mg A Betst Reenus fig. I IVB 14 NA hes ong ong casuso ong

a) Let the velocity at A be $v_{A}$ .

Applying conservation of mechanical energy between the highest point of the track and point A:

$mgh=\frac{1}{2}mv_{A}^{2}$

UA 2gh

From figure 2:

The resultant of normal force by the track and weight provides the necessary centripetal force:

$N_{A}-mg=\frac{mv_{A}^{2}}{R}$

$\Rightarrow N_{A}=\frac{m(\sqrt{2gh})^{2}}{R}+mg$

$\Rightarrow N_{A}=\frac{m*2gh}{R}+mg=mg(1+\frac{2h}{R})$

Therefore, the force of the track on the block at point A is $N_{A}=mg(1+\frac{2h}{R})$ [answer]

b) Let the speed at point B is $v_{B}$ .

The height of point B from the lowest point is $h'=R(1-cos45^{\circ})$ .

Applying conservation of mechanical energy between A and B:

$\frac{1}{2}mv_{A}^{2}=mgh'+\frac{1}{2}mv_{B}^{2}$

$\Rightarrow v_{A}^{2}=2gR(1-cos45^{\circ})+v_{B}^{2}$

$\Rightarrow v_{B}^{2}=v_{A}^{2}-2gR(1-cos45^{\circ})$

#ub= 2gh – 2gR(1 - cos45°)

Now, the resultant of normal force by the track and component of weight along the radius provides the necessary centripetal force (refer figure 3):

$N_{B}-mg\cos45^{\circ}=\frac{mv_{B}^{2}}{R}$

$\Rightarrow N_{B}=\frac{m(2gh-2gR(1-cos45^{\circ}))}{R}+mg\cos45^{\circ}$

$\Rightarrow N_{B}=\frac{mg(2h-R(2-3\cos45^{\circ}))}{R}$

Therefore, the force of the track on the block at point B is $N_{B}=\frac{mg(2h-R(2-3\cos45^{\circ}))}{R}$ [answer]

c) The block leaves the track at point B with the speed $v_{B}=\sqrt{2gh-2gR(1-cos45^{\circ})}$ [answer]

d) The vertical component of speed $v_{B}$ is $v_{B}\sin45^{\circ}$ directed upward.

The horizontal component of speed $v_{B}$ is $v_{B}\cos45^{\circ}$ directed towards right.

When the block hits the ground the vertical displacement is $s_{y}=-h' = -R(1-cos45^{\circ})$

Let the time taken by the block to hit the ground is 't' from B.

Applying kinematic equation along vertical:

$s_{y}=v_{B}\sin45^{\circ}*t-\frac{1}{2}gt^{2}$

$\Rightarrow -R(1-cos45^{\circ})=(\sqrt{2gh-2gR(1-cos45^{\circ})})\sin45^{\circ}*t-\frac{1}{2}gt^{2}$

$\Rightarrow gt^{2}-2(\sqrt{2gh-2gR(1-cos45^{\circ})})\sin45^{\circ}*t-2R(1-cos45^{\circ})=0$

$\Rightarrow t=\frac{2(\sqrt{2gh-2gR(1-cos45^{\circ})})\sin45^{\circ}\pm \sqrt{(2(\sqrt{2gh-2gR(1-cos45^{\circ})})\sin45^{\circ})^{2}-4*g*-2R(1-cos45^{\circ})}}{2g}$

$\Rightarrow t=\frac{2(\sqrt{2gh-2gR(1-cos45^{\circ})})\sin45^{\circ}\pm \sqrt{(4(2gh-2gR(1-cos45^{\circ}))\sin^{2}45^{\circ})-4*g*-2R(1-cos45^{\circ})}}{2g}$

$\Rightarrow t=\frac{2(\sqrt{2gh-2gR(1-cos45^{\circ})})\sin45^{\circ}\pm \sqrt{2(2gh-2gR(1-cos45^{\circ}))+8gR(1-cos45^{\circ})}}{2g}$

$\Rightarrow t=\frac{2(\sqrt{2gh-2gR(1-cos45^{\circ})})\sin45^{\circ}\pm \sqrt{(4gh+4gR(1-cos45^{\circ}))}}{2g}$

$\Rightarrow t=\frac{(\sqrt{gh-gR(1-cos45^{\circ})})\pm \sqrt{(gh+gR(1-cos45^{\circ}))}}{g}$

$\Rightarrow t=\frac{(\sqrt{h-R(1-cos45^{\circ})})+ \sqrt{(h+R(1-cos45^{\circ}))}}{\sqrt{g}}$

Now, the horizontal distance travelled from B is

$S=v_{B}\cos45^{\circ} *t$

  $=\frac{(\sqrt{h-R(1-cos45^{\circ})})+ \sqrt{(h+R(1-cos45^{\circ}))}}{\sqrt{g}}*\left (\sqrt{2gh-2gR(1-cos45^{\circ})}*cos45^{\circ} \right )$

  $=\left (\sqrt{h-R(1-cos45^{\circ})}+ \sqrt{(h+R(1-cos45^{\circ}))} \right )*\left (\sqrt{h-R(1-cos45^{\circ})} \right )$

Therefore, the distance of the point where the block hits the ground from A is

$S+Rsin45^{\circ}$

$=\left (h-R(1-cos45^{\circ})+ \sqrt{h^{2}-R^{2}(1-cos45^{\circ})^{2}} \right )+Rsin45^{\circ}$[answer]