Question

2 doors are uniform and identical. Door A rotates about an axis through its left edge, and door B rotates about an axis through its center. the same force is applied perpendicular to each door at its right edge and force remains perpendicular as door turns. no other force affects the rotation of either door. starting from rest door A rotates through a certain angle in 3.00s. how long does it take door b (also from rest) to rotate through the same angle?

Answer 1

1/2 of A:)

Answer 2

the answer is suppose to be 4x A?

Answer 3

Okaay....? 4A?
\(\tau_1=Fd=I_1 \alpha_1 \)...(1), where \(\alpha= \frac{2 \theta}{t^2}\)
Similarly,\(\tau_2=F(\frac{d}{2})...(2)=I_2 \alpha_2 \)
Now, from here,
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
I sub into (1) and (2) and then take the ratio (1):(2),
\(\frac{Fd}{F(\frac{d}{2})} =\frac{\frac{(ML^2)(2 \theta)}{3t^2_1}}{\frac{(ML^2)(2 \theta)}{12t^2_2}}\)
And this gives
\(t_2 = \sqrt 2 t_1\)...Did I miss something?

Answer 4

\(t_2 = t_1/ \sqrt 2\)

Answer 5

Sorry, miscalc'ed. \(t_2=\frac{\sqrt2 t_1}{2}\). Are you sure it's 4A.?

Answer 6

vincent -lyon.fr is right..the answer will be t2=t1/√2..

Answer 7

maybe answer key is wrong? :s

Answer 8

i dont understand where the 3 and 12 come from in the ratio equation..

Answer 9

\(I_A=ML^2/3\) whereas \(I_B=ML^2/12\) : moments of inertia are different.
\(\Gamma_A=FL\) whereas \(\Gamma_B=FL/2\)

Angular momentum is \(\Gamma/I\)
Hence \(\alpha_A=\Large \frac{3F}{ML}\) and \(\alpha_B=\Large \frac{6F}{ML}\)

Rotated angle is \(\theta_A = \frac12 \alpha_A t_A^2\) and \(\theta_B = \frac12 \alpha_B t_B^2\)

Answer 10

The 3 and 12 are obtained from referring to the moments of inertia in the link above. :) Note that different things have different moment of inertias based on different axes.