AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A student skates up a ramp that makes an angle 30° with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed 𝑣₀ and wants to turn around over a semicircular path xyz of radius 𝑅 during which he/she reaches a maximum height ℎ (at point y) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then (g is the acceleration due to gravity)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$v_{0}^{2} - 2 gh = \frac{1}{2} gR$

the centripetal force required at points x and z is zero

$v_{0}^{2} - 2 gh = \frac{\sqrt{3}}{2} gR$

the centripetal force required is maximum at points x and z

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

By the energy conservation (ME) between bottom point and point Y

$\begin{array}{l} \frac{1}{2} {mv}_{0}^{2} = mgh + \frac{1}{2} {mv}_{1}^{2} \\ ∴ v_{1}^{2} = v_{0}^{2} - 2 gh \dots (i) \end{array}$

Now at point Y the centripetal force provided by the component of mg

$\begin{array}{l} ∴ mgsin 30^{\circ} = \frac{{mv}_{1}^{2}}{R} \\ ∴ v_{1}^{2} = \frac{gR}{2} \\ ∴ from (i) \\ \frac{g R}{2} = v_{0}^{2} - 2 g h \end{array}$

As gravitation force is not providing centripetal force at x and z that's why maximum force is applied on x and z for circular motion.

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!