A disc of radius r=0.1m is rolled from a point A on a track as shown in the figure. The part AB of the track is a semi-circle of radius R in a vertical plane. The disc rolls without sliding and leaves contact with the track at its highest point B. Flying through the air it strikes the ground at point C. The velocity of the center of mass of the disc makes an angle of 300 below the horizontal at the time of striking the ground. At the same instant, velocity of the topmost point P of the disc is found to be 6 m/s. The value of R( in m) is (Take g=10 m/s2 ).

Question

A disc of radius r=0.1m is rolled from a point A on a track as shown in the figure. The part AB of the track is a semi-circle of radius R in a vertical plane. The disc rolls without sliding and leaves contact with the track at its highest point B. Flying through the air it strikes the ground at point C. The velocity of the center of mass of the disc makes an angle of 300 below the horizontal at the time of striking the ground. At the same instant, velocity of the topmost point P of the disc is found to be 6 m/s. The value of R( in m) is (Take g=10 m/s2 ).

Infinity Learn · Accepted Answer

Since the disc was rolling, the velocity of its top point at the instant of leaving the track was zero. Its means horizontal component of com of disc $v_{H} = ω r$ .
When the disc is in air $v_{H}$ and $ω$ both do not change. Hence the horizontal component of the velocity of the top point P of the disc at every instant is zero and the vertical component of the velocity of the point p is equal to the vertical component of velocity of the CM of disc.
$\Rightarrow \sqrt{2 g h} = \sqrt{2. g .2 (R - 0.1)} = 6 \Rightarrow 4 \times 10 (R - 0.1) = 36 R = 1 m$

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