Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Rolling Motion›Question 32

JEE Main 2024 (1 Feb, Shift 2) · Rolling Motion

Rolling Motion question from JEE Main 2024 (1 Feb, Shift 2)

Q32JEE Main 2024 · 1 FEB · SHIFT 2Medium

Rolling Motion

A disc of radius RR and mass MM is rolling horizontally without slipping with speed vv. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is:

Figure

Show answer and solution

Answer: (a)

  1. If the incline gives rolling (as the key assumes): KE of disc =12Mv2(1+12)=34Mv2= \frac{1}{2}Mv^2\left(1+\frac{1}{2}\right) = \frac{3}{4}Mv^2.
  2. Mgh=34Mv2⇒h=3v24gMgh = \frac{3}{4}Mv^2 \Rightarrow h = \dfrac{3v^2}{4g}.
  3. So the answer is (a) 34v2g\frac{3}{4}\frac{v^2}{g}.
Report a mistake

More questions on Rolling Motion