Question

Figure shows the roller coaster track. Each car will start from rest at point $A$ and roll with negligible friction. It is important that there should be at least some small positive normal force exerted by the track on the car at all points, otherwise the car would leave the track. With the above fact, the minimum safe value for the radius of curvature at point B is $(g=10m/s^2)$

• A. $20m$
• B. $10m$
• C. $40m$
• D. $25m$

Answer 1

The correct option is A $20m$

As car is in rest position intially, so $v=0m/s$.

At point B,
$mg-N=\frac{m{v}^2}{R}$
where, $N=$ normal force, $g$= acceleration due to gravity, $v$= speed of car
$\therefore N=mg-\frac{m{v}^2}{R}$ ......(1)
At point $B$,
Potential energy of car is zero.
and at point $A$,
friction is zero and $v=0m/s$ (as given)
Hence, $TM{E}_A=PE+KE$ ....(2)
Where, $KE=0$ because $v=0m/s$

and at point $B$,
$TM{E}_B=PE+KE$.....(3)
Here, $PE$ will be zero.
By equation (2) and equation (3)
$TM{E}_A=TM{E}_B$
$mg\left(10\right)=\frac{1}{2}m{v}^2$
$\to v^2=20g$
on putting this value at equation (1)
$N=mg-\frac{20mg}{R}$
For contact with track, $N>0$
Hence, $mg-\frac{20mg}{R}>0$
$mg>\frac{20mg}{R}$
So, $R>20m$
So the radius of curvature at point B should be more then $20m$ for contact with path at point $B$.
So, option (A) matches correctly.