Question

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).

Answer 1

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