SATHEE: System of Particles and Rotational Motion

System of Particles and Rotational Motion - Result Question 70

77. A ring of mass $m$ and radius $r$ rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $ω$ . Its kinetic energy is

[1988]

(a) $\frac{1}{2} m r^{2} ω^{2}$

(b) $m r ω^{2}$

(d) $\frac{1}{2} m r ω^{2}$

Topic 5: Rolling Motion

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Answer:

Correct Answer: 77. (a)

Solution:

(a) Kinetic energy $= \frac{1}{2} I ω^{2}$

and for ring $I = m r^{2}$

Hence, $K E = \frac{1}{2} m r^{2} ω^{2}$

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System of Particles and Rotational Motion - Result Question 70

77. A ring of mass m and radius r rotates about an axis passing through its centre and perpendicular to its plane with angular velocity ω. Its kinetic energy is

Answer:

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77. A ring of mass $m$ and radius $r$ rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $ω$ . Its kinetic energy is