Physics speed vs time graph?
The figure gives angular speed versus time for a thin rod that rotates around one end. (a) What is the magnitude of the rod's angular acceleration? (b) At t = 4.0 s, the rod has a rotational kinetic energy of 5.08 J. What is its kinetic energy at t = 0?
I just need answer for b
the answer for a is 1.5
The figure gives angular speed versus time for a thin rod that rotates around one end. (a) What is the magnitude of the rod's angular acceleration? (b) At t = 4.0 s, the rod has a rotational kinetic energy of 5.08 J. What is its kinetic energy at t = 0?
I just need answer for b
the answer for a is 1.5
1 Answer
See the answer below....
Explanation:
- The graph cuts two point
#color(red)((2,1);(4,4)# . So, the equation of the straight line of the graph is
#(y-4)/(x-4)=(4-1)/(4-2)#
#=>3x-2y=4#
#y=underbrace(color(red)(3/2)) x-2# The gradient that is the value of angular acceleration is
#color(red)(1.5)" rad/s".#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let, at
#t=0# , the angular velocity is#omega_1# and at#t=4# , the angular velocity is#omega_2#
The formula to determine the rotational kinetic energy is
#color(red)("E"_k=1/2 m omega^2r^2" r" "="radius"# I can now write,
#1/2 m omega_2^2r^2=5.08#
#=>1/2 m xx4^2xxr^2=5.08" ""as "omega_2=4 #
#=>mr^2=127/200# In the equation, I determined, taking
#x(t)=0# , we get#y(omega_1)=-2# .The kinetic energy at
#t=0# is#=>"E"_k=color(green)(1/2xxmr^2 xx omega_1^2)=1/2xx127/200xx(-2)^2=color(red)(2.54" J"# Hope it helps...
Thank you..