A car starts from rest at height H and moves | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a car to complete a vertical loop of radius $r$,the minimum velocity at the top of the loop must be $v_{top} = \sqrt{gr}$.Using the principle

Vedclass · Accepted Answer

$H/h = 5$

Vedclass · Answer

(D) For a car to complete a vertical loop of radius $r$,the minimum velocity at the top of the loop must be $v_{top} = \sqrt{gr}$.Using the principle of conservation of energy between the starting point (height $H$) and the top of the loop (height $2r$):$mgH = mg(2r) + \frac{1}{2}mv_{top}^2$$mgH = 2mgr + \frac{1}{2}m(gr)$$H = 2r + 0.5r = 2.5r = \frac{5r}{2}$.The height $h$ is the height of the top of the loop above the ground level of the loop,which is $2r$. However,the problem defines $h$ as the height of the loop above the starting point of the loop track. Given the diagram,the loop starts at ground level,so the top of the loop is at height $2r$. Thus $h = 2r$.From $H = 2.5r$ and $h = 2r$,we have $r = h/2$.Substituting $r$ into the equation for $H$: $H = 2.5(h/2) = 1.25h$.Wait,re-evaluating the standard problem: If the car starts from height $H$ and enters a loop of radius $r$,the condition to complete the loop is $H = 2.5r$. The top of the loop is at height $2r$. If $h$ is defined as the height of the top of the loop from the ground,$h = 2r$. Then $H/h = 2.5r / 2r = 1.25$. Given the options provided $(2, 3, 4, 5)$,let's re-examine the definition of $h$. If $h$ is the height of the loop above the ground,$h=2r$. If the question implies $H = h + 2r$ where $h$ is the height above the loop,then $H = 2.5r$ and $h = 0.5r$,so $H/h = 2.5/0.5 = 5$.

$A$ car starts from rest at height $H$ and moves along a track to complete a vertical loop of radius $r$. If the car just completes the loop,find the ratio $H/h$,where $h$ is the height of the loop above the starting point of the loop track.

Similar Questions

$A$ ball attached to one end of a string swings in a vertical plane such that its acceleration at point $A$ (extreme position) is equal to its acceleration at point $B$ (mean position). The angle $\theta$ is :-

The mass of the bob of a simple pendulum of length $L$ is $m$. If the bob is released from its horizontal position,then the speed of the bob and the tension in the thread at the lowest position of the bob will be respectively:

$A$ particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance $h$ below the highest point,such that

$A$ stone of mass $1 \ kg$ is tied to a string $2 \ m$ long and is rotated at a constant speed of $40 \ ms^{-1}$ in a vertical circle. The ratio of the tension at the top and the bottom is [Take $g = 10 \ ms^{-2}$].

Vedclass Test Series

Exam Paper Generator

Online Exam Module