Question

A truck at rest has one door fully open as shown in figure
a)
1 The truck accelerates forward at a constant rate A, and the door begins to swing

shut. The door is uniform and solid, with height h, width w and mass M. Neglect air

resistance.

(a) Find the instantaneous angular velocity of the door about its hinges when it has

swung through 90◦

.

(b) Find the horizontal force on the door when it has swung through 90◦



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Answer 1

Heyy!! Here is your answer !

Track Door
1)The door's centre of mass falls a distance w/2 as it gains rotational kinetic energy.

Moment of Inertia about hinge of door =( Ml^2)/3
By conservation of Mechanical Energy,

Let the required angular velocity be (k).

1/2 I k^2 = M A w/2

=> k = (3A/w)^1/2.

2) The equation of motion is
F(h) - MA = M w/2 *k^2
=>F(h) = 5/2 MA

There fore , Horizontal force required is 5*2MA.


Hope, you understand my answer and it may helps you!

Answer 2

A)

The center of mass of the door is at its geometric center, h/2 from the top or bottom edge, and w/2 from the left or right edge. The moment of inertia is, using the parallel axis theorem, (1/12)Mw² + M(w/2)² = (1/3)Mw².

B)

The truck is uniformly accelerating horizontally, so in a frame moving with the truck, it appears as though there is a horizontal component to gravity with a value of -A. Conservation of energy then gives: MAw/2 = 1/2 Iω² or ω = (3A/w)½.

Thanks