hollow sphere moment of inertia
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Answers:we start with newton's second law the sum of the forces acting on the sphere will equal ma where a is the linear acceleration the forces are the component of gravity down the plane and the force of friction up the plane this gives us mg sin (theta) - u mg cos(theta) = ma where u is the coeff of friction a=gsin(theta) - u g cos(theta) Now, consider the torque on the rolling sphere. The force of friction generates a torque on the sphere; the lever arm of this force is the radius of the sphere, so we have torque = f R since f= u mg cos(theta), the torque is: torque = umg cos(theta) R now, we use the fact that torque = I A where I is the moment of inertia (for a sphere this is 2/5 mR^2) and A is the angular acceleration A is related to linear acceleration via a= R A or A=a/R so we have: torque = u mg cos(theta) R = I A = 2/5 mR^2(a/R) so this gives us u g cos(theta)=2/5 a now return to g sin(theta) - u gcos(theta) = a substitute 2/5 a for u g cos(theta) to get g sin(theta) - 2/5 a = a g sin(theta) = 7/5 a or a=5/7 g sin(theta)
Answers:definitely not (b) two objects falling from the same height will fall according to the newtonian equations of motion. all are mass independant definitely not (c) scales measure mass or weight. weight = mass x g. since mass is the same and g is the same, the balance would read the same for both objects leaves (a) so why would the balls roll differently? they are the same shape. same size. but one has all it's mass around the outside of the sphere and the other is uniform..... so here's the answer..... moment of inertia for a rotating body is a measure of it's resistence to rotate. at a given applied torque, an object with a higher moment of inertia will be more resistent to rotate. ie.. it will accelerate slower. the moment of inertia of a hollow sphere = 2 m r / 3 the moment of inertia of a solid sphere = 2 m r / 5 since m is the same and so is r.... and since 2/3 > 2/5, the hollow sphere will have a higher moment of inertia and therefore will be more resistent to rotate than the solid sphere for a given torque. in this case, that torque is due to the weight of the object via normal force and friction. the torque = the force of friction. so the solid sphere will accelerate faster down the incline ******** update ********** I see someone gave me a thumbs down. so maybe they want more details? t = I where t = torque = force x moment arm I = moment of inertia = angular acceleration so = t / I for a ball rolling down an incline, torque is the friction imparted by the surface = x Fn (coeff of friction x normal force). but Fn = m x a x cos (m x a = weight of object acting straight down = angle of incline so ma = Fn... draw a pic if you need).. so t = m a makes = t / I = m a / I for hollow sphere, = m a / (2 m r /3) = 3 a / (2 r ) for solid sphere, = m a / (2 m r /5) = 5 a / (2 r ) solid / hollow = 5 a / (2 r ) / 3 a / (2 r ) = 5/3 solid = 5/3 hollow ie, the acceleration of the solid sphere = 5/3 x the acceleration of the hollow sphere.... keep in mind, the sphere is not sliding down the plane. it is rolling....
Answers:A good experiment/project involving energy conservation and angular momentum/moment of inertia is to have the students build a well (like a somewhat u-shaped section of a roller coaster track), acquire several common objects with different moments of inertia (solid sphere, hollow sphere, solid disk, hollow disk, ...) , and then test how they act differently in the well. For instance, start them from the same height and see what height they return to and how long each takes to reach it's max height once again. What they will find is that they all return to their original height due to energy conservation, but that they will take different times to do so depending on there moment of inertia.
Answers:_________________________________________________ Outer radius = R =0.43 m inner radius =r=0.17 m friction force = f The torque due to frictional force= T=R*f but torque T =moment of inertia I*angular acceleration(a!) moment of inertia= I = (1/2)*m*(R^2 + r^2) the acceleration of the cylinders center of mass =a angular acceleration a!=a / R R*f = I*a!=Ia/R f = Ia /R^2 Applied force =F=49.8 N mass=m=2.56 kg Applying Newton's second law, ma= F - f ma = F - I a/R^2 ma + Ia/R^2 = F a= F / [m +I/R^2 ] substituting, I =(1/2)m[R^2+r^2] a = F / m [1+(1/2)[1 +( r/R)^2 ] a = 2 F / m [3+( r/R)^2 ] a =12.3265 m/s^2 the acceleration of the cylinders center of mass is 12.3265 m/s^2 _____________________________________________________________________