The depicted wheel rolls without slipping, if omega is 100 RPM, and r = 50 cm, w
ID: 1645793 • Letter: T
Question
The depicted wheel rolls without slipping, if omega is 100 RPM, and r = 50 cm, what is the overall velocity at points A, B, and O (magnitude and direction)? A block with a mass of 1.0 kg and an initial speed of vi = 2.0 m/s slides on a frictionless horizontal surface. The block encounters a spring that is in its equilibrium position, and compresses it until the block comes to rest momentarily. Find the maximum compression of the spring, assuming its force constant is 5 N/cm. Indicate the direction of the force The baseball of mass = 0.145 kg is directed towards the player with velocity 40 m/s. He hits it back and it travels with velocity 50 m/s in the opposite direction. (a) What is the impulse imparted to the ball? (b) If the ball and bat are in contact for 1.5 ms, what is average experienced force?Explanation / Answer
Given
the wheel with radius r = 0.5 m, W = 100 RPM = 100*2pi/60 rad/s = 10.472 rad/s
as the wheel is rolling without slipping so v = r*W = 0.5*10.472 = 5.236 m/s at "O"
V_B = v - W*r , v_A = v+Wr
at B ,v_B=0 ==> v= W*r ===>v_A = 2*V
the velocity at B is zero , because of the friction balances the wheel without slipping
and velocity at A (top) is double the velocity v ===> v_A = 2*5.236 m/s = 10.472 m/s
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Given
block mass m = 1.0 kg, vi=2.0 m/s , vf = 0 m/s
spring constant or force constant is k = 5 N/cm = 500 N/m
here the block will come to rest after the spring compresses through a distance x that is
as the block compressed the spring the elastic potential enrgy will develop in spring so that the velocity of the block finally becomes zero so
the total kineticenergy of the block = elastic potential energy of the spring
0.5*m(vf^2-vi^2 ) = 0.5*k*x^2
x^2 = m(vi^2 )/k
x = sqrt(m(vf^2-vi^2 )/k)
x = sqrt(1(2^2)/500) m
x = 0.936 m
the spring force F = -kx = -(500*0.936 ) N = -468 N
the negative sign indicates the direction of the force is to the left , in the opposite direction.
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base ball mass m = 0.145 kg, initial velocity towards the player vi = 40 m/s
velocity after hit is vf = -50 m/s
a)
impulse imparted to the ball is m(vf-vi) = 0.145(-50-40) = -13.05 kg m/s
b) time t = 1.5 ms , the average force experienced is i = F*t
F = i/t = -13.05/(1.5*10^-3) N
F = -8700 N
the force is F = 8700 N