TS EAMCET 4-Aug-2021 Shift 1 Question Paper

The given situation is shown belowMass of small disc, m=500g
Radius of disc, R=5cm
Initial kinetic energy of the disc will be zero i.e.
Ki=0
Final kinetic energy of the disc
Kf=‌

1

2

mv2
where, v is velocity of disc at bottom.
According to conservation of energy.
Total energy at point A= Total energy at point O
⇒Ui+(K‌rot ‌)‌initial ‌+Ki=Uf+(K‌rot ‌)‌final ‌+Kf
⇒‌‌mgh+0+0=0+‌

1

2

Iω2+‌

1

2

mv2
⇒‌‌mgh=‌

1

2

Iω2+‌

1

2

mv2
‌ Since, ‌ω=‌

v

R

‌ and ‌I=‌

1

2

mR2
∴‌‌mgh=‌

1

2

×‌

1

2

mR2×‌

v

R

+‌

1

2

mv2
⇒‌‌v=√‌

4

3

gh
⇒‌‌v∝√h
Hence, speed of centre of mass of disc at bottom depends on height of inclined plane.