A long, uniform rod of mass 20.00 kg is pivoted at its top end. It is lifted a s

Question

A long, uniform rod of mass 20.00 kg is pivoted at its top end. It is lifted a small angle and let go. When it reaches the very bottom of its swing, it is moving with angular speed ?o , and it has collision with a block of mass 3.000 kg that was originally at rest on a frictionless surface. After the collision the rod is seen to continue spinning in the same direction at 17.66 rad/s. As a result of the collision, the 3.000-kg block moves to the right along the frictionless surface, climbs the ramp and leaves at an angle of 16.96? when it is 1.840 m high. The block flies as a projectile and when it reaches the very top of its flight it has an elastic collision with a second, larger block, which was at rest on top of a 2.210-m high slide. As a result of the collision, the second block slides down the frictionless curved slide, after which it crosses a very long, flat rough section, and then hits a spring. The spring compresses 0.3000 m from its equilibrium position. The spring-constant of the spring is 1200 N/m. The block attaches to the spring and the system oscillates in simple harmonic motion with a period of 0.3630 s.

1. Find the velocities of blocks A and B immediately after they collide.

2. Find the initial angular speed of the rod, Wo

top of flight nnally attest 0.40 m ABy at rest 16 2.21 m SHM with T-0.363 s 20.0 kg 1.84 m -0.280 no friction 213.4 m max compression 3.00 kg ncialy at rest ong rough section 0.300 m .

Explanation / Answer

time period T = 2*pi8sqrt(m/k)

given T = 0.363

K = 1200

0.363 = 2*3.14*sqrt(m/1200)

mB = 4 kg

the initial energy of block B at height 2.21 m

0.5*mB*vB'^2 + mB*g*2.21 = 0.5*4*vB'^2 + (4*9.8*2.21)

E1 = 2vB^2 + 86.632

work done Wf = uk*mB*g*x = 0.28*4*9.8*13.4 = 147.0784 J

energy stored in spring E2 = 0.5*k*dx^2 = 0.5*1200*0.3*0.3 = 54 J

from work energy theorem

E1 - E2 = W

2vB'^2 + 86.632 - 54 = 147.0784

vB' = 7.56 m/s

for block A from 1.84 m to 2.21m

dy = 2.21-1.84 = 0.37 m

initial velocity = voy = v*sin16.96

final velocity at 2.21m = vfy = 0

2*g*dy = vfy^2 - voy^2

2*9.8*0.37 = voy^2

voy = 2.69 m/s

vo = 9.23 m/s

speed of block A at 2.21m

vA = vA*cos16

vA = 8.83 m/s

at 2.21 m

initial momentum = final momentum

mA*vA + mB*vB = mA*vA' + mB*vB'

(3*8.83)+(4*0) = (3*vA')+(4*7.56)

vA' = -1.25 m/s

-----------------------

(2)

energy of A at 1.84m

0.5*mA*vo^2 + mA*g*1.84 = (0.5*3*9.23*9.23)+(3*9.8*1.84)

EA = 181.88535

energy of A immediately after collision with rod = KE = 0.5*mA*v^2

initial energy = final enery at 1.84m

0.5*3*v^2 = 181.88535

v = 11.01 m/s

initial angular momentum = final angular momentum

I*wo + = I*w + mA*v*l

I = (1/3)*M*l^2 = 1.067 kg m^2

(1.067*wo) = (1.067*17.66)+(3*11.01*0.4)

wo = 30.04 rad/s <---answer

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