Your shopping cart is empty!
Explore Related Concepts


Best Results From Yahoo Answers
From Yahoo Answers
Question:HELP!!! What is the linear acceleration of the center of mass of a solid sphere?
A solid sphere (radius R, mass M) rolls down an incline as shown in the figure**. Its linear
acceleration of the center of mass is what?
**the figure could not be pasted onto yahoo, so here is the link:
http://www.scribd.com/doc/18103177/SerwayPhysicChapter11
It is problem number 20 (just scroll down a bit), and the figure is on the right hand side. Thanks.
The answer is (5/7)(g)(sin ). How was this derived?
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: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)
Question:Two spheres look identical and have the same mass.One is hollow and other is solid.Which which method would determine which is which?
a. roll them down and incline
b. drop them from the same height
c. weigh them on a scale
Thank you:>
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: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....
Question:What's a good physics energy or energy conservation activity for 11th or 12th grade physics? Visual and/or hands on would be best. Thanks!
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 ushaped 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:A good experiment/project involving energy conservation and angular momentum/moment of inertia is to have the students build a well (like a somewhat ushaped 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.
Question:A 2.56 kg hollow cylinder with inner radius
0.17 m and outer radius 0:43 m rolls with
out slipping when it is pulled by a horizontal
string with a force of 49.8 N
Its moment of inertia about the center of
mass is (1/2)*m*(R^2out + R^2in)
What is the acceleration of the cylinders center of mass?
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 _____________________________________________________________________
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 _____________________________________________________________________