Rigid Structure Figure 12-27 shows a rigid structure consisting of a circular ho

Question

Rigid Structure Figure 12-27 shows a rigid structure consisting of a circular hoop of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 1.0 s.

Figure 12-27
(a) Assuming R = 0.80 m and m = 2.0 kg, calculate the structure's rotational inertia about the axis of rotation.
kg·m2
(b) Calculate the structure's rotational momentum about the axis of rotation.
kg·m2/s



DIAGRAM

http://img715.imageshack.us/i/1239u.gif/

Explanation / Answer

a)from the diagram rotational inertia about the axis of rotation                      I =mR2/3+mR2/3+mR2+mR2/2+mR2----1 this is because axis is at the end of two rods therefore moment of inertia of two rods about this axis is                        = mR2/3+mR2/3 and at distance R from another rod then moment ofinertia of this rod about this axis is mR2 and this axis is passing through one rod moment of inertia iszero and this axis passing through the tangent of the circular ring in the plane of paper then moment of inertia is                        I = mR2/2+mR2 add all these moment of inertia GIVEN m = 2.0kg R = 0.8 meters and period of rotation T = 1.0s plug the values in the above equation 1 and get moment ofinertia     I = mR^2[ ( 1/3) + (1/3) + 1 +(1/2) + (1) ]       = (2)(0.8)2 [3.167]       =  4.053 kg.m^2 b) The angular momentum L = I = ANGULAR SPEED = 2/TIME PERIOD                                =(2/1.0)rad/sec                                = 6.283 rad/sec Therefore the angular momentum            L = (1.45)(3.925) = 25.465 kgm^2/s a)from the diagram rotational inertia about the axis of rotation                      I =mR2/3+mR2/3+mR2+mR2/2+mR2----1 this is because axis is at the end of two rods therefore moment of inertia of two rods about this axis is                        = mR2/3+mR2/3 and at distance R from another rod then moment ofinertia of this rod about this axis is mR2 and this axis is passing through one rod moment of inertia iszero and this axis passing through the tangent of the circular ring in the plane of paper then moment of inertia is                        I = mR2/2+mR2 add all these moment of inertia GIVEN m = 2.0kg R = 0.8 meters and period of rotation T = 1.0s plug the values in the above equation 1 and get moment ofinertia     I = mR^2[ ( 1/3) + (1/3) + 1 +(1/2) + (1) ]       = (2)(0.8)2 [3.167]       =  4.053 kg.m^2 b) The angular momentum L = I = ANGULAR SPEED = 2/TIME PERIOD                                =(2/1.0)rad/sec                                = 6.283 rad/sec Therefore the angular momentum            L = (1.45)(3.925) = 25.465 kgm^2/s a)from the diagram rotational inertia about the axis of rotation                      I =mR2/3+mR2/3+mR2+mR2/2+mR2----1 this is because axis is at the end of two rods therefore moment of inertia of two rods about this axis is                        = mR2/3+mR2/3 and at distance R from another rod then moment ofinertia of this rod about this axis is mR2 and this axis is passing through one rod moment of inertia iszero and this axis passing through the tangent of the circular ring in the plane of paper then moment of inertia is                        I = mR2/2+mR2 add all these moment of inertia GIVEN m = 2.0kg R = 0.8 meters and period of rotation T = 1.0s plug the values in the above equation 1 and get moment ofinertia     I = mR^2[ ( 1/3) + (1/3) + 1 +(1/2) + (1) ]       = (2)(0.8)2 [3.167]       =  4.053 kg.m^2 b) The angular momentum L = I = ANGULAR SPEED = 2/TIME PERIOD                                =(2/1.0)rad/sec                                = 6.283 rad/sec Therefore the angular momentum            L = (1.45)(3.925) = 25.465 kgm^2/s a)from the diagram rotational inertia about the axis of rotation                      I =mR2/3+mR2/3+mR2+mR2/2+mR2----1 this is because axis is at the end of two rods therefore moment of inertia of two rods about this axis is                        = mR2/3+mR2/3 and at distance R from another rod then moment ofinertia of this rod about this axis is mR2 and this axis is passing through one rod moment of inertia iszero and this axis passing through the tangent of the circular ring in the plane of paper then moment of inertia is                        I = mR2/2+mR2 add all these moment of inertia GIVEN m = 2.0kg R = 0.8 meters and period of rotation T = 1.0s plug the values in the above equation 1 and get moment ofinertia     I = mR^2[ ( 1/3) + (1/3) + 1 +(1/2) + (1) ]       = (2)(0.8)2 [3.167]       =  4.053 kg.m^2 b) The angular momentum L = I = ANGULAR SPEED = 2/TIME PERIOD                                =(2/1.0)rad/sec                                = 6.283 rad/sec Therefore the angular momentum            L = (1.45)(3.925) = 25.465 kgm^2/s

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