A circular disc of radius r is rotating about its axis o with a uniform angular velocity
Answers
Explanation:
Answer: 2 R ω sin (θ/2)
Explanation:
The radius of the circular disc = R, about its axis O
Uniform angular velocity = ω rad/sec
Referring to the figure attached below, we can say that the velocity at point A and point B will be “r * ω”. So, we can consider that both the velocity vectors “Va” & “Vb” will meet at an angle “θ”.
Therefore,
The magnitude of relative velocity of point A relative to point B on the disc is,
| Vab |
= |Va - Vb|
= √[Va² +Vb² – (2*Va*Vb*cosθ)]
= √[(R² ω²) + (R² ω²) – {2* (R²ω²) cosθ}]
= R ω √[2* (1 – cosθ)]
= R * ω * √2 * √[2 * {(1 – cosθ)/2}]
= R * ω * √2 * √2 * √[(1 – cosθ)/2]
using half angle identity, sin (θ/2) = ± √[(1 – cosθ)/2]
= 2 R ω sin (θ/2)
Explanation:
Answer: 2 R ω sin (θ/2)
Explanation:
The radius of the circular disc = R, about its axis O
Uniform angular velocity = ω rad/sec
Referring to the figure attached below, we can say that the velocity at point A and point B will be “r * ω”. So, we can consider that both the velocity vectors “Va” & “Vb” will meet at an angle “θ”.
Therefore,
The magnitude of relative velocity of point A relative to point B on the disc is,
| Vab |
= |Va - Vb|
= √[Va² +Vb² – (2*Va*Vb*cosθ)]
= √[(R² ω²) + (R² ω²) – {2* (R²ω²) cosθ}]
= R ω √[2* (1 – cosθ)]
= R * ω * √2 * √[2 * {(1 – cosθ)/2}]
= R * ω * √2 * √2 * √[(1 – cosθ)/2]
using half angle identity, sin (θ/2) = ± √[(1 – cosθ)/2]
= 2 R ω sin (θ/2)