In: Physics
In the figure, a small block of mass m = 0.121 kg slides down a frictionless surface from an initial height of h = 0.850 m and then sticks to a uniform vertical rod of mass M = 0.879 kg and length L = 1.83 m. The rod pivots about point O through an angle θ before momentarily stopping. Find θ (in degrees).
given
m = 0.121 kg
M = 0.879 kg
L = 1.83 m
speed of m before touchinng the rod, v = sqrt(2*g*h)
= sqrt(2*9.8*0.85)
= 4.08 m/s
Let w is the angular speed of rod just after the collision.
Apply conservation of angular momentum
m*v*L = (M*L^2/3 + m*L^2)*w
==> w = m*v*L/(M*L^2/3 + m*L^2)
= 0.121*4.08*1.83/(0.879*1.83^2/3 + 0.121*1.83^2)
= 0.651 rad/s
kinetic energy just after the collision, KE =
0.5*I*w^2
= 0.5*(M*L^2/3 + m*L^2)*w^2
= 0.5*(0.879*1.83^2/3 + 0.121*1.83^2)*0.651^2
= 0.2938 J
let theta is the angle made by the rod with vertical before momentarily stopping.
height raised by center of mass of rod, h = (L/2)*(1 - cOs(theta))
height raised by block, h2 = L*(1 - cOs(theta))
Apply conservation of Energy
KEi = PEf
0.2938 = M*g*h1 + m*g*h2
0.2938 = M*g*(L/2)*(1 - cOs(theta)) + m*g*L*(1 - cOs(theta))
0.2938 = 0.879*9.8*(1.83/2)*(1- cos(theta)) + 0.121*9.8*1.83*(1 - cos(theta))
0.2938 = 10.052*(1 - cos(theta))
1 - cos(theta) = 0.2938/10.052
1 - cos(theta) = 0.03
cos(theta) = 1 - 0.03
cos(theta) = 0.97
theta = cos^-1(0.97)
= 14.07 degrees <<<<<<<<<<<<-------------Answer