A light rod of length 1.00 m rotates about an axis perpendicular to its length a

Question

A light rod of length 1.00 m rotates about an axis perpendicular to its length and passing through its center. Two particles of masses 4.00 kg and 3.00 kg are connected to the end of the rod. Neglecting the mass of the rod, what is the system's rotational kinetic energy when its angular speed is 2.50 rad/s? A metal rod that is 4.00 m long and 0.50 cm^2 in cross-sectional area is elongated to 0.20 cm under a tensile force of 5000 N. What is the elastic modulus for this metal? A Circular steel rod 2.00 m long and 5.00 m diameter is elongated to 0.75 cm when a tensile force of 250 N is applied to the rod. What is the Young's Modulus of the steel rod?

Explanation / Answer

Given

   masses of particles m1 = 4 kg, m2 = 3 kg

   length of the rod l = 1m

the moment of inertia of the system of masses is

   I = m1*r1^2 +m2*r2^2

here r1= r2 = 0.5 m

   = 4*0.5^2 +3*0.5^2 kg m2
   = 1.75 kg m2

and the angular speed w = 2.5 rad/s

now the rotational kinetic energy is k.e = 0.5*I*W^2

       k.e = 0.5*1.75*2.5^2 J

       k.e = 5.46875 J

2.

   Given
length of the rod is l = 4 m
elangation in the length is Dl = 0.2 m

tensile force is F = 5000 N

area is 0.5*10^-4 m2

nwo the elastic modulus is y = F*l /A*Dl

               = 5000*4/(0.5*10^-4*0.2)

               = 2000000000 N/m2

3. diameter of the rod is d = 5m , radius is r = 2.5 m , area is a = pi*r^2 = pi*2.5^2 = 19.635 m2

elangation is Dl = 0.75cm = 0.0075 m

tensile force F = 250 N

   youngs modulus is y = F*l /A*Dl

               = 250*2/(19.635*0.0075)

               = 3395.2975129445717 N/m2

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