Consider the following figure in the horizonal plane (ie no gravity). Assume small motion - that...

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Consider the following figure in the horizonal plane (ie no gravity). Assume "small" motion - that is, you can assume that the angle the rocker arm rotates remains small, 0 = 0 is when the rocker arm is horizontal and corresponds to the unstretched position of the spring. The quantity Jois the mass moment of inertia about point O of the rocker arm (ie the horizontal bar), k is the stiffness of the (massless) linear spring that is fixed at one end, and M is the external moment imposed by the cam on the system. This moment is produced by the contact force generated by the cam at the left end of the rocker arm. Syntax notes: when entering your responses, use J[0] for Jo (the subscript is the number zero). Use theta for 0. You can use the loupe/magnifying glass to check that what you entered is what you intended. The mass m is rigidly attached to the rocker arm by a weighless rod. We can tre at the rocker Jo and mass m as one rigid body (let's call it the rocker-arm assembly). Not clear from the figure, but please treat mass m as if it is sitting on the bar at b. Then, the moment of inertia of the rocker arm assembly (bar + mass m) about point O is given by JA Positive is defined in the counterclockwise direction. When the bar rotates by a small angle 0, then the moment of the spring about point O is given by (syntax: use theta to denote the angle, you need to account for the direction of the moment - ie think about + or - in front of your math expression). Mspring= B The assembly is rotating about pivot O. Below, we assume that the signs for the moments will be included into their expressions when you write them out (later). Netwon's second law of motion gives the eqation of motion as O Jo d e(t) dt d e(t) OJA dt2 d e(t) OJA dt = - M ae(t) O Jo dt a = M -=M + Mspring =M + Mspring Why does it make sense to sum the moments about the pivot point O? O By taking the moments about point O, the reaction forces at O don't need to be found because their moments are zero. O Taking the moments about point O means that we can ignore gravity. O None of the other responses are correct. O It doesn't make sense to take the moments about O, we should write F= ma for the centre of mass of the bar, and F=ma for the mass m O Taking the moments about point O guarantees that the small angle assumption is valid. Hence, the equation of motion of the system, from Newton's second law of motion, is given by Hint: your answer should be in terms of the original given variables: Jo, M,m, k, a, b, and and its derivatives. For derivatives, you have to write the explicit time dependence, otherwise Maple will evaluate it to zero. Meaning: for de you need to write diffitheta(t),t) and not diff(theta,t). You can write a double derivative as diff(theta(t) tt). The equation of motion is given by Now suppose that we consider the same system but in the vertical plane - that is, we can no longer ignore gravity. Furthermore, assume that the rocker arm centre mass is at the pivot point O. How does this now change our equation of motion? O It does not change our equation of motion. All the other responses are incorrect. O We need to add another moment, Mg mgb to the sum of moments. O We need to add another moment, Mg mgb to the sum of moments. = Consider the following figure in the horizonal plane (ie no gravity). Assume "small" motion - that is, you can assume that the angle the rocker arm rotates remains small, 0 = 0 is when the rocker arm is horizontal and corresponds to the unstretched position of the spring. The quantity Jois the mass moment of inertia about point O of the rocker arm (ie the horizontal bar), k is the stiffness of the (massless) linear spring that is fixed at one end, and M is the external moment imposed by the cam on the system. This moment is produced by the contact force generated by the cam at the left end of the rocker arm. Syntax notes: when entering your responses, use J[0] for Jo (the subscript is the number zero). Use theta for 0. You can use the loupe/magnifying glass to check that what you entered is what you intended. The mass m is rigidly attached to the rocker arm by a weighless rod. We can tre at the rocker Jo and mass m as one rigid body (let's call it the rocker-arm assembly). Not clear from the figure, but please treat mass m as if it is sitting on the bar at b. Then, the moment of inertia of the rocker arm assembly (bar + mass m) about point O is given by JA Positive is defined in the counterclockwise direction. When the bar rotates by a small angle 0, then the moment of the spring about point O is given by (syntax: use theta to denote the angle, you need to account for the direction of the moment - ie think about + or - in front of your math expression). Mspring= B The assembly is rotating about pivot O. Below, we assume that the signs for the moments will be included into their expressions when you write them out (later). Netwon's second law of motion gives the eqation of motion as O Jo d e(t) dt d e(t) OJA dt2 d e(t) OJA dt = - M ae(t) O Jo dt a = M -=M + Mspring =M + Mspring Why does it make sense to sum the moments about the pivot point O? O By taking the moments about point O, the reaction forces at O don't need to be found because their moments are zero. O Taking the moments about point O means that we can ignore gravity. O None of the other responses are correct. O It doesn't make sense to take the moments about O, we should write F= ma for the centre of mass of the bar, and F=ma for the mass m O Taking the moments about point O guarantees that the small angle assumption is valid. Hence, the equation of motion of the system, from Newton's second law of motion, is given by Hint: your answer should be in terms of the original given variables: Jo, M,m, k, a, b, and and its derivatives. For derivatives, you have to write the explicit time dependence, otherwise Maple will evaluate it to zero. Meaning: for de you need to write diffitheta(t),t) and not diff(theta,t). You can write a double derivative as diff(theta(t) tt). The equation of motion is given by Now suppose that we consider the same system but in the vertical plane - that is, we can no longer ignore gravity. Furthermore, assume that the rocker arm centre mass is at the pivot point O. How does this now change our equation of motion? O It does not change our equation of motion. All the other responses are incorrect. O We need to add another moment, Mg mgb to the sum of moments. O We need to add another moment, Mg mgb to the sum of moments. = Consider the following figure in the horizonal plane (ie no gravity). Assume "small" motion - that is, you can assume that the angle the rocker arm rotates remains small, 0 = 0 is when the rocker arm is horizontal and corresponds to the unstretched position of the spring. The quantity Jois the mass moment of inertia about point O of the rocker arm (ie the horizontal bar), k is the stiffness of the (massless) linear spring that is fixed at one end, and M is the external moment imposed by the cam on the system. This moment is produced by the contact force generated by the cam at the left end of the rocker arm. Syntax notes: when entering your responses, use J[0] for Jo (the subscript is the number zero). Use theta for 0. You can use the loupe/magnifying glass to check that what you entered is what you intended. The mass m is rigidly attached to the rocker arm by a weighless rod. We can tre at the rocker Jo and mass m as one rigid body (let's call it the rocker-arm assembly). Not clear from the figure, but please treat mass m as if it is sitting on the bar at b. Then, the moment of inertia of the rocker arm assembly (bar + mass m) about point O is given by JA Positive is defined in the counterclockwise direction. When the bar rotates by a small angle 0, then the moment of the spring about point O is given by (syntax: use theta to denote the angle, you need to account for the direction of the moment - ie think about + or - in front of your math expression). Mspring= B The assembly is rotating about pivot O. Below, we assume that the signs for the moments will be included into their expressions when you write them out (later). Netwon's second law of motion gives the eqation of motion as O Jo d e(t) dt d e(t) OJA dt2 d e(t) OJA dt = - M ae(t) O Jo dt a = M -=M + Mspring =M + Mspring Why does it make sense to sum the moments about the pivot point O? O By taking the moments about point O, the reaction forces at O don't need to be found because their moments are zero. O Taking the moments about point O means that we can ignore gravity. O None of the other responses are correct. O It doesn't make sense to take the moments about O, we should write F= ma for the centre of mass of the bar, and F=ma for the mass m O Taking the moments about point O guarantees that the small angle assumption is valid. Hence, the equation of motion of the system, from Newton's second law of motion, is given by Hint: your answer should be in terms of the original given variables: Jo, M,m, k, a, b, and and its derivatives. For derivatives, you have to write the explicit time dependence, otherwise Maple will evaluate it to zero. Meaning: for de you need to write diffitheta(t),t) and not diff(theta,t). You can write a double derivative as diff(theta(t) tt). The equation of motion is given by Now suppose that we consider the same system but in the vertical plane - that is, we can no longer ignore gravity. Furthermore, assume that the rocker arm centre mass is at the pivot point O. How does this now change our equation of motion? O It does not change our equation of motion. All the other responses are incorrect. O We need to add another moment, Mg mgb to the sum of moments. O We need to add another moment, Mg mgb to the sum of moments. =

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