The kinetic energy of a circular disc rotating with a speed of 60 r.p.m. about an axis passing through a point on its circumference and perpendicular to its plane is (mass of circular disc =5 kg, radius of operatornamedisc=1 m ) approximately.

Question

The kinetic energy of a circular disc rotating with a speed of 60 r.p.m. about an axis passing through a point on its circumference and perpendicular to its plane is (mass of circular disc =5 kg, radius of operatornamedisc=1 m ) approximately. (A) 170 J (B) 160 J (C) 150 J (D) 140 J

Tardigrade Admin · Accepted Answer

The rotational kinetic energy is given by KE =(1/2) I ω2=(1/2) × (3/2) mr 2 4 π2 f2 where, m= mass, r= radius and f= frequency ⇒ KE =3 mr 2 × π2 f 2=3 × 5 × 1 × 10 × 1 ⇒ KE =150 J

Q. The kinetic energy of a circular disc rotating with a speed of $60$ r.p.m. about an axis passing through a point on its circumference and perpendicular to its plane is (mass of circular disc $= 5 k g$ , radius of $disc = 1 m$ ) approximately.

170 J

160 J

150 J

140 J

The rotational kinetic energy is given by
$K E = \frac{1}{2} I ω^{2} = \frac{1}{2} \times \frac{3}{2} m r^{2} 4 π^{2} f^{2}$
where, $m =$ mass, $r =$ radius and $f =$ frequency
$\Rightarrow K E = 3 m r^{2} \times π^{2} f^{2} = 3 \times 5 \times 1 \times 10 \times 1$
$\Rightarrow K E = 150 J$

Q. The kinetic energy of a circular disc rotating with a speed of 60 r.p.m. about an axis passing through a point on its circumference and perpendicular to its plane is (mass of circular disc =5kg, radius of disc=1m ) approximately.