Answer:
a. 1.5rad/s
b. 0.40J
Explanation:
a. Angular acceleration is obtained by determining the gradient of the straight line.
-The graph perfectly cuts two points[4,4], [2,1]:
[tex]Gradient=\frac{\bigtriangleup y}{\bigtriangleup x}\\\\=\frac{4-1}{4-2}\\\\=1.5[/tex]
#The line's gradient is 1.5
Hence, the rod's angular acceleration is 1.5rad/s
b. Let at t=0, the angular velocity be [tex]w_1[/tex] and [tex]w_2[/tex] at [tex]t=4[/tex].
#Rotational kinetic energy is determined using the formula:
[tex]E_k=\frac{1}{2}m\omega^2r^2, r=radius[/tex]
#Equate the expression to the energy value given:
[tex]E_k=\frac{1}{2}m\omega_2^2r^2\\\\\frac{1}{2}m\omega_2^2r^2=1.60J, \ \omega_2=4\\\\\frac{1}{2}m\times 4^2\times r^2=1.60\\\\mr^2=0.20\\\\\# Taking \ x(t)=0->y(\omega_1)=-2\\\\\therefore E_k=0.5mr^2\times \omega_1^2, \ \ \#t=0\\\\E_k=0.5\times 0.2\times (-2)^2\\\\=0.40[/tex]
Hence, the kinetic energy at t=o is 0.40J
