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When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it falls. If the distance to the floor is 78 cm and for rotation less than 1 rev, what are the (a) smallest and (b) largest angular speeds that cause the toast to hit and then topple to be butter-side down? Assume free-fall acceleration to be equal to 9.81 m/s2.

Respuesta :

Answer:

a) 3.94 rad/s

b) 11.84 rad/s

Explanation:

Given:

Distance, s = 78 cm = 0.78 m

Now the time taken, t

we know

[tex]s = ut +\frac{1}{2}gt^2[/tex]

where,

s  = distance

u = initial speed

g = acceleration due to gravity

since it is a free fall. thus, u = 0

thus, we get

[tex]0.78 = 0\times t +\frac{1}{2}9.8\times t^2[/tex]

or

[tex]t=\sqrt{\frac{2\times 0.78}{9.8}}=0.398 s[/tex]

a) now, the smallest angle will be 1/4 of the revolution so as to fall on the butter side i.e 90° or ([tex]\frac{\pi}{2}[/tex])

also,

[tex]angular\ speed\ (\omega) = \frac{Change\ in\ angle}{time}[/tex]

thus, we have

[tex]angular\ speed\ (\omega) = \frac{\frac{\pi}{2}}{0.398} = 3.94rad/s[/tex]

b)now, the largest angle will be 3/4 of the revolution so as to fall on the butter side i.e 270° or ([tex]\frac{3\pi}{2}[/tex]) contributing to the largest angular speed

[tex]angular\ speed\ (\omega) = \frac{Change\ in\ angle}{time}[/tex]

thus, we have

[tex]angular\ speed\ (\omega) = \frac{\frac{3\pi}{2}}{0.398} =11.84rad/s[/tex]

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