Losing contact

Classical Mechanics Level 3

The figure above shows a smooth track in a vertical plane, consisting of two circular arcs A B AB and B C BC of same radius r = 2 m . r=2\text{ m}. The common tangent to the arcs, B D , BD, is horizontal. A small block placed on the track at a height h h (in centimeters) above B D BD leaves the track at exactly the same depth below. Find h . h.

10 40 80 20

View wiki
Prashant Kr by Prashant Kr , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

When the centrifugal force exceeds the component of gravitational force, block will leave the track at point L. O is the center of the lower curve. In falling through 2h under gravity, the square of the velocity of the b l o c k a t L = V 2 = 2 ∗ g ∗ 2 h ∴ C e n t r i f u g a l a c c e l e r a t i o n = V 2 r = 4 g h r . . . . ( 1 ) α is the angle between vertical and a radial line OL. L is at a distance of h from BD. ∴ C o s α = r − h r . Component of gravitational acceleration opposing c e n t r i f u g a l a c c e l e r a t i o n = g ∗ C o s α = r − h r ∗ g . . . . . ( 2 ) S i n c e m a s s i s s a m e ( 1 ) = ( 2 ) ⟹ 4 g h r = r − h r ∗ g . ∴ 4 h = r − h ⟹ 5 h = 2 a n d h = . 4 m = 40 c m \text{When the centrifugal force exceeds the component of gravitational force,}\\\text{block will leave the track at point L. O is the center of the lower curve.}\\ \text{In falling through 2h under gravity, the square of the velocity of the}\\ block ~ at ~ L = V^2=2*g*2h\\\therefore Centrifugal ~acceleration=\dfrac{V^2}{r}=\dfrac{4gh}{r}....(1)\\\alpha \text{ is the angle between vertical and a radial line OL.} \\ \text{L is at a distance of h from BD. }~~~~~\therefore Cos\alpha=\dfrac{r-h}{r}.\\ \text{Component of gravitational acceleration opposing } \\centrifugal ~~acceleration =g*Cos\alpha=\dfrac{r-h}{r}*g.....(2)\\Since~ mass~ is ~same ~(1) = (2)\implies \dfrac{4gh}{r}=\dfrac{r-h}{r}*g.\\\therefore 4h=r-h \implies 5h=2~and~h=.4m=\boxed{\color{#3D99F6}{\Large 40 cm}}
Sorry, have taken r in place of R.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

The terminology "centrifugal acceleration" is wrong in this case.

Atomsky Jahid - 5 years ago

Log in to reply

May I know why and what is the right term ? Thanks.

Niranjan Khanderia - 5 years ago

Log in to reply

If you set centrifugal force and gravitational force equal to each other, they will cancel each other out because they work in opposite direction. And, when the net force becomes zero an object will move in a straight line, not in a circular path. (Newton's first law!) The correct interpretation is "The gravitational force is providing the centripetal force". Reference: Page 155, University Physics, Young and Freedman, 13th edition.

Atomsky Jahid - 4 years, 12 months ago

Log in to reply

@Atomsky Jahid – You are correct. It is the gravity that provides centripetal force. But it IS a centripetal force.

Niranjan Khanderia - 4 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...