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 79 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.

Given:

Distance, S = 79 cm

= 0.79 m

Acceleration due to gravity, g = 9.81 m/s^2

Theta = <1 rev

= < 2pi rad

A.

Using equation of motion,

S = vo * t + 1/2 * g * t^2

Initially at rest, vo = 0 m/s

t = sqrt ((0.79 * 2) / 9.81)

= 0.40 s

The smallest and largest angle that can cause the toast to hit and then topple to be butter-side down is guven by the range;

0.75 rev > theta > 0.25 rev

A.

Angular velocity = angular displacement/time

= (0.25 * 2pi) / 0.4

= 3.92 rad/s

B.

Angular velocity = angular displacement/time

= (0.75 * 2pi) / 0.4

= 11.74 rad/s

a) 5.61 rad/s

b) 16.83 rad/s

Explanation:

Distance to the floor = 79cm = 0.79m

Rotation is <1 revolution

The toast will rotate at an angular speed that is constant while it falls. The toast will also be falling with constant acceleration due to gravity. Using equation of motion, which is

S = ut + 1/2gt²

S = 0 + 1/2gt²

S = gt², where, S = d

0.79 = 9.81t²

t² = 0.79/9.81

t² = 0.081

t = 0.28s

As the toast is accidentally pushed, it rotates as it falls. It will be landing on its edges when and if it hits the ground. The smallest angle here would then be 1/4 of the revolution. This is also the smallest angular speed.

ω (min) = ΔΦ revolution / Δt

ω (min) = 1/4 * 2π / 0.28

ω (min) = 0.5π/0.28

ω (min) = 5.61 rad/s

Since 1/4 of revolution is the minimum angle, the remaining (3/4), is the maximum angle. Thus

ω (max) = 3/4 * 2π / 0.28

ω (max) = 0.75 * 2π / 0.28

ω (max) = 16.83 rad/s