A uniform rod of mass M=15 kg and length 2l=2 m is pinned vertically to a wall a

Question

A uniform rod of mass M=15 kg and length 2l=2 m is pinned vertically to a wall at one end and is free to rotate. Initially, the rod is balanced vertically upwards as shown in the figure above. A small piece of sticky-puddy, mass m=1 kg and with initial speed v0 = 10m/s, comes flying towards the top end of the rod (as shown) and collides with it. The collision is perfectly inelastic. Ignore any friction between the rod and the pin.

(i) Determine the angular speed of the puddy-rod system immediately after the collision.
(ii) Determine the angular speed of the puddy-rod system when it is pointed vertically downwards.

Explanation / Answer

i) since it is perfectly inelastic collision

va=vb= maua+mbub / ma+mb

here a is the ball and b is the top most point of rod, ua=10 m/s, ub=0 (rod is at rest)

va=vb= 10+0 / (1+15)

va=vb=0.625 m/s

angular speed=v/ length= 0.625/2=0.3125 rad/s

ii) conservation of energy

0.5*M*vtop2+ M*4l*g = 0.5*M*vbottom2 {M is the combined Mass of both}

sqrt(2*(0.5*0.6252 + 4*9.8))=vbottom

vbottom=8.876 m/s

angular speed at bottom =vbottom/length= 8.876/2=4.438 rad/s

iii) let the mass puddy system be a and unknown block be b (let say its mass is mb=m)

initial and final velocity for collision be u and v respectively then,

ma=16, ua=8.876 m/s, ub=0,

to find va use energy conservation just like above and put vtop=0 and find vbottom=va

va=8.854 m/s

writing both energy and momentum conservation

16*8.876+m*0 = 16*8.854 + m*vb

16*8.8762=16*8.8542+m*vb2

solving these two we get

mb=0.01985 kg = 19.85 g

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.