A rectrangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. when relesed, it slips off the table in a certain short time T=0.01s, remaining essentiallly horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

  • Option 1)

    0.5

  • Option 2)

    0.3

  • Option 3)

    0.02

  • Option 4)

    0.28

Answers (1)

Apply Angular impulse= change in angular momentum

Tdt=\Delta \L

T=mg\frac{\l}{2}

Tdt=\Delta \L

\left ( mg \right )\frac{\l}{2}\times 0.01=\left ( \frac{m\l^{2}}{3} \right )w

w=\frac{3 \times 10 \times 0.01}{2 \times 0.3}=0.05 rad/s

now t=time taken by rad. to hit the ground

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1 sec

so \Theta =angle rotate by rad

\Theta = wt

\Theta = 0.5 rad


Option 1)

0.5

Option 2)

0.3

Option 3)

0.02

Option 4)

0.28

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