Answer:
a.3.99rad/s
b.11.9604rad/s
Explanation:
Let,[tex]d=76cm=0.76m[/tex] be distance from the counter to the floor and rotation of less than 1rev. [tex]t[/tex] 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:
[tex]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[/tex]
#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.
[tex]\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[/tex]
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.
[tex]\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[/tex]
Hence, the largest angular speed is 11.9604rad/s
