Question

A meter stick is held vertically with one end on the floor and is allowed to fall . The speed of the other end when it hits the floor assuming that the end at the floor does not slip(g=9.8)

Answer 1

L = length of meterstick = 1 m

h = height of center of mass of meterstick = L/2 = 1/2 = 0.5 m

m = mass of meterstick

w = angular speed just before the stick hits the floor

I = moment of inertia of meterstick = mL²/3

Using conservation of energy

potential energy = rotational kinetic energy

mgh = (0.5) I w²

m g (L/2) = (0.5) ( mL²/3) (v/L)²

mgL = mv²/3

v = sqrt(3gL)

inserting the values

v = sqrt(3 x 9.8 x 1)

v = 5.42 m/s

Answer 2

Answer:

The speed of the other end when it hits the floor is 5.42 m/s.

Explanation:

Let,

$L$ = length of meterstick =$1 m$;

$h$ = height of center of mass of meterstick = $\frac{L}{2}=\frac{1}{2}=0.5m$ ;

$m$ = mass of meterstick;

$w$ = angular speed just before the stick hits the floor;

$I$ = moment of inertia of rod/meter stick = $\frac{mL^2}{3}$ ;

Potential of center of mass of the stick = $mgh =mg\frac{L}{2}$

Rotational kinetic energy = $\frac{1}{2} Iw^2$

Using the conservation of energy,

potential energy = rotational kinetic energy

$mgh = (0.5) I w^2$

$\implies m g \frac{L}{2} = (0.5)\frac{mL^2}{3} (\frac{v}{L})^2$

$\implies mgL = mv^2/3$

$\implies v = \sqrt {(3gL)$

$\implies v = \sqrt {(3 *9.8 * 1)$

$\implies v = 5.42 m/s$

The speed of the other end when it hits the floor is 5.42 m/s.

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