Calculate the following!!!!!!!! The disk rolls without slipping. If the force ac

Question

Calculate the following!!!!!!!!

The disk rolls without slipping. If the force acts through a distance of 5 m. what is the angular velocity in rad/s of the disk if the disk starts from rest? I = (1/2) mr: 2 4 6 8 10 The 2 kg mass falls 300 mm from rest. If the spring of stiffness 400 N/m is initially relaxed and the mass is at rest, determine the final speed in m/s of the mass. Neglect friction. 1 2 3 4 5. A 40 kg block is pulled along a frictionless surface by a force varying as shown. If the block starts at rest, what is its speed in m/s after it has been pulled 5 m? 2.5 5 5 2 10 A 2 m long uniform rod of mass 4 kg is pivoted about one end. It is released from rest in the horizontal position. What is its angular velocity when it reaches vertical. The moment of inertia is (1/3) mL2. 15 I0 5 5 e. 10 A body w ith a moment of inertia of 24 kg m: is freely rotating at an angular velocity of 30 rad/s. A portion of the body slips increasing the moment of inertia to 72 kg m:. What is its new angular velocity? 3 10 30 45 90

Explanation / Answer

32) Torque T = I*alpha = r*F = 1*100 = 100

I = 0.5*m*r^2 = =0.5*20*1*1 = 10 kg*m^2

alpha = T/I = 100/10 = 10

apply w = sqrt(2*r*alpha*S) = sqrt(2*1*10*5) = 10 rad/s

so answer is e 10

33) 0.5*k*x^2 = 0.5*m*v^2

v = sqrt(k/m)*x = sqrt(400/2)*0.300 = 4.24

So the answer is d) 4

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34) F =m*a

a = F/m =200/40 = 5 m/s^2

apply v = sqrt(2*a*s) = sqrt(2*5*5) = 5*sqrt(2)

So the answer is c) 5*sqrt(2)

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35) 0.5*I*w^2= m*g*h = 4*9.81 = 39.24

I= (1/3)*4*2*2 = 5.33 kg*m^2

then w = sqrt(39.24/(.5*5.33)) = sqrt(15)

so the answer is a) sqrt(15)

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36) I1 = 24 kg*m^2

w1 = 30 rad/s

I2 = 72 kg*m^2

w2 = ?

I1*w1 = I2*w2

w2 = I1*w1/I2= 24*30/72 = 10 rad/s

so the answer is b) 10

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