Question

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 76 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

Answer 1

Answer:

a.3.99rad/s

b.11.9604rad/s

Explanation:

Let,$d=76cm=0.76m$ be distance from the counter to the floor and rotation of less than 1rev. $t$ is the time it takes to hit the floor.

#The toast rotates at a constant angular speed as it falls, with a constant gravitational acceleration.

#Using Kinematic equations:

$d=v_it+0.5gt^2\\d=0+0.5gt^2\\\therefore t=\sqrt{\frac{2d}{g}}=\sqrt{\frac{2\times 0.76}{9.8}}\\\\=0.394s$

#The lands on one of the edges if it topples to be butter-side down. Thus the smallest angle is 0.25rev which corresponds to the smallest angular velocity.

$\omega_m_i_n=\frac{\bigtriangleup \theta}{\bigtriangleup t}\\\\\omega_m_i_n=\frac{0.25rev}{\bigtriangleup t}=\frac{0.25\times2\pi}{\bigtriangleup t}\\\\\omega_m_i_n=\frac{0.5\pi}{0.394}=3.99rad/s$

Hence, the smallest angular speed is 3.99rad/s

b. #The lands on one of the edges if it topples to be butter-side down. Thus the smallest angle is 0.75rev which corresponds to the largest angular velocity.

$\omega_m_i_n=\frac{\bigtriangleup \theta}{\bigtriangleup t}\\\\\omega_m_i_n=\frac{0.75rev}{\bigtriangleup t}=\frac{0.75\times2\pi}{\bigtriangleup t}\\\\\omega_m_i_n=\frac{1.5\pi}{0.394}=11.9604rad/s$

Hence, the largest angular speed is 11.9604rad/s