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]:
![Gradient=\frac{\bigtriangleup y}{\bigtriangleup x}\\\\=\frac{4-1}{4-2}\\\\=1.5](https://tex.z-dn.net/?f=Gradient%3D%5Cfrac%7B%5Cbigtriangleup%20y%7D%7B%5Cbigtriangleup%20x%7D%5C%5C%5C%5C%3D%5Cfrac%7B4-1%7D%7B4-2%7D%5C%5C%5C%5C%3D1.5)
#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
and
at
.
#Rotational kinetic energy is determined using the formula:
![E_k=\frac{1}{2}m\omega^2r^2, r=radius](https://tex.z-dn.net/?f=E_k%3D%5Cfrac%7B1%7D%7B2%7Dm%5Comega%5E2r%5E2%2C%20r%3Dradius)
#Equate the expression to the energy value given:
![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](https://tex.z-dn.net/?f=E_k%3D%5Cfrac%7B1%7D%7B2%7Dm%5Comega_2%5E2r%5E2%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7Dm%5Comega_2%5E2r%5E2%3D1.60J%2C%20%5C%20%5Comega_2%3D4%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7Dm%5Ctimes%204%5E2%5Ctimes%20r%5E2%3D1.60%5C%5C%5C%5Cmr%5E2%3D0.20%5C%5C%5C%5C%5C%23%20Taking%20%5C%20x%28t%29%3D0-%3Ey%28%5Comega_1%29%3D-2%5C%5C%5C%5C%5Ctherefore%20E_k%3D0.5mr%5E2%5Ctimes%20%5Comega_1%5E2%2C%20%5C%20%5C%20%5C%23t%3D0%5C%5C%5C%5CE_k%3D0.5%5Ctimes%200.2%5Ctimes%20%28-2%29%5E2%5C%5C%5C%5C%3D0.40)
Hence, the kinetic energy at t=o is 0.40J