Question

a circular disc of mass 1 kg and radius 2m is revolving about an axis passing through its centre and perpendicular to its plane the disc makes 100 revolutions per minute calculate it's angular momentum and kinetic energy​

Answer 1

AnsWer :-

✏ We're given that,a circular disc of mass (m) 1 kg and radius (r) 2m is revolving about an axis passing its centre and perpendicular to its plane the disc makes 100 revolutions per minute i.e Angular velocity (ω) is 100 rev/min.

✏ We need to find the angular momentum (L) and rotational kinetic energy (K.E)

✣ S T E P S : ✣

✒First we have to convert the angular velocity from rev/min to rev/sec.

✒Then we need to find the moment of inertia (I) of the disc revolving about an axis passing through its centre and perpendicular to its plane.

✒After finding moment of inertia of the disc we can calculate the angular momentum easily.

✒At last we can find the rotational kinetic energy.

✪ C A L C U L A T I O N : ✪

☯ Angular velocity (ω) : ☯

↠ω = 100 rev/min

↠ω = 100 × (2π/60)

↠ω = 100 × (π/30)

↠ω = 100π/30

↠ω = (100 × 3.14)/30

↠ω = (10 × 3.14)/3

↠ω = 31.4/3

↠ω ≈ 10.47 rev/sec

∴ The angular velocity of the disc is 10.47 rev/s.

☯ Moment of Inertia (I) : ☯

⇢I = (MR²)/2

⇢I = (1 × [2]²)/2

⇢I = 4/2

⇢I = 2 kg.m²

∴ The moment of inertia (I) of the disc revolving about an axis passing through its centre and perpendicular to its plane is 2 kg.m².

☯ Angular momentum (L) : ☯

⇸L = Iω

⇸L = 2 × 10.47

⇸L = 2 × 10.47

⇸L = 20.94

⇸L ≈ 21 kg.m².s-¹

∴ The angular momentum (L) of the disc is 21 kg.m².s-¹ (approx).

☯ Rotational kinetic energy : ☯

→ K.E = ½ × I × ω²

→ K.E = ½ × 2 × (10.47)²

→ K.E = ½ × 2 × 109.7

→ K.E = 109.7 J

∴ Rotational kinetic energy of the disc is 109.7 J.

Answer 2

Answer:

A circular disc of mass 1 kg and radius 2m is revolving about an axis passing through its center and perpendicular to its plane the disc makes 100 revolutions per minute.  

• Angular momentum ( L ) =  $20.9\ kgm^{2} /s$

• Rotational kinetic energy $(K . E_{\text {rotational }})= 109.55\ J$

Explanation:

 

Given,  

Mass of the disc   (M) = 1 kg  

The radius of the disc (R)  =  2 m  

Angular velocity $(\omega)=100\ rev/min=\frac{100 \times 2 \pi}{60}=\frac{10}{3} \pi = 10.466 \ rev/sec$  

For a circular disc, the moment of inertia is given by,

$\mathrm{I}=\frac{M R^{2}}{2}$    

  $=\frac{1}{2} \times 1 \times 2^{2} = 2 kgm^{2}$

Angular momentum can be expressed as,

$L = I$ω

   = $2\times \frac{10\pi }{3} =\frac{20\pi }{3}$ =$20.9 kgm^{2} /s$

Rotational kinetic energy is given by the equation,

$K . E_{\text {rotational }}=\frac{1}{2} I \omega^{2}$

                     =  $\frac{1}{2} \times2 \times(\frac{10}{3} \pi)^{2}$

                     =  $\frac{100}{9} \times(3.14)^{2}$

                     = 109.55 J

Ans :

Angular momentum ( L ) =  $20.9\ kgm^{2} /s$

Rotational kinetic energy ($K . E_{\text {rotational }})= 109.55\ J$