Respuesta :
Answer:
Yes, I get 17 rad/s², too.
Note that the assumption of constant angular acceleration is really, really, terrible. A valid answer to this question (i.e., one that does not assume constant angular acceleration) involves differential equations. But if you do assume constant angular acceleration, this is quite straightforward. Use constant-acceleration kinematics:
Δθ = ω_i Δt + ½α (Δt)²
You know the pencil moves through an angle of π/2 radians. The initial angular velocity is zero. You already found the angular acceleration, and you want Δt.
Δt = âš[ 2 Δθ / α ] = âš[ 2 (Ď€/2 rad) / 17 rad/s² ] = 0.34 s
This is the same calculation oldprof makes, but his treatment of the pencil as a point mass rather than a uniform rod has thrown his angular acceleration off.
Answer:
[tex]\alpha = \frac{3gsin10}{2L}[/tex]
Explanation:
As we know by Newton's law
[tex]\tau = I\alpha[/tex]
here torque is due to the weight of the pencil which will act on the center of the pencil
so here we will have
[tex]\tau = mg\frac{L}{2}sin\theta[/tex]
now we know that the moment of inertia of the pencil is given as
[tex]I = \frac{mL^2}{3}[/tex]
now we will have
[tex]mg\frac{L}{2}sin\theta = \frac{mL^2}{3}\alpha[/tex]
[tex]\alpha = \frac{3g}{2L}sin\theta[/tex]
for 10 degree angle we have
[tex]\alpha = \frac{3gsin10}{2L}[/tex]
