A a disk, a solid sphere, and a hollow spherical shell of the same mass and radius are released?
Rex "old teacher" is absolutely correct. The fastest order is solid sphere, disc, hollow sphere, this is because of the inertia solid sphere = 0.7mv^2 = mgh disc = 0.75mv^2 = mgh hollow sphere = 0.833mv^2 = mgh solid sphere velocity = sq-root((gh)/0.7) disc velocity = sq-root((gh)/0.75) hollow sphere velocity = sq-root((gh)/0.833)
A solid sphere, a hollow sphere, a solid cylinder, and a hollow cylinder are released from the top of an inclined plane. Which object arrives first?
We need to assume that each object has uniform density and that they all roll without slipping. Perhaps surprisingly size, mass, density height don’t matter. Definemass [math]=m[/math]initial height [math]=h[/math]gravity [math]=g[/math]final velocity [math]=v[/math]radius [math]=r[/math]angular velocity [math]= \omega=\frac{v}{r}[/math]moment of inertia [math]=I=kmr^2[/math] (where [math]k[/math] is a measure how close to the edge the mass is, on average)Then by conservation of energy,[math]mgh=\frac12 m v^2 +\frac12 I \omega^2 [/math][math]2gh= v^2 + k v^2 [/math][math]v=\sqrt{\dfrac{2gh}{1+k}}[/math]So to maximize [math]v[/math] we need to minimize [math]k[/math] - in other words we want the mass to be concentrated as close to the centre as possible. This is intuitive, we don’t want to ‘waste’ energy spinning up the object - we want to concentrate on maximizing its speed.Therefore the slowest object will be the hollow cylinder (all the mass is at the edge) and the fastest will be the solid sphere.
A spherical shell, a solid sphere and solid cylinder are released from rest at the top of an inclined plane,?
As the three start with the same total energy TE = mgh, each will end up with the same kinetic energy at the bottom of the ramp. But, this is important, the distribution of that KE between angular and linear kinetic energy will differ according to the distribution of that mass, i.e., their shapes. k = 2/5 for the solid sphere, = 2/3 for the hollow one, and = 1/2 for the solid cylinder; each having r radius of gyration. So KE = 1/2 mV^2 (1 + k) is the total energy at the bottom of the ramp for each one. As 1 + k indexes 100% of the KE, we have: 7/5 ~ 100% and 5/5//7/5 = 1/(1 + k) = 5/7 of the solid sphere's energy is linear. 5/3 ~ 100% and 1//5/3 = 3/5 of the hollow sphere energy is linear. 3/2 ~ 100%. And by similar means 2/3 of the cylinder's KE is linear. So the rank order, first to last to the bottom will be: solid sphere, solid cyl., hollow sphere. ANS.
What is total kinetic energy when a solid sphere of mass 10 kg and diameter of 1 m rolls without slipping with uniform speed 5 m/sec on a horizontal surface?
KK (total) =(1/2)*(7/5)*m*v² = 0,7 * m*v² = 0.7*10*5² = 0.7*250 = 175 Joules.Source for the image above: http://ask.learncbse.in/t/a-soli....
Four objects - a hoop, a solid cylinder, a solid sphere, and a thin, spherical shell?
Four objects - a hoop, a solid cylinder, a solid sphere, and a thin, spherical shell - each has a mass of 4.80 kg and a radius of 0.212 m. IMG- http://www.webassign.net/sercp8/p8-44.gif (a) Find the moment of inertia for each object as it rotates about the axes shown in the table above. hoop kg·m2 solid cylinder kg·m2 solid sphere kg·m2 thin, spherical shell kg·m2 (b) Suppose each object is rolled down a ramp. Rank the translational speed of each object from highest to lowest. 1. solid cylinder > thin spherical > solid sphere > hoop 2. solid sphere > solid cylinder > thin spherical > hoop 3. hoop > solid cylinder > solid sphere > thin spherical 4. thin spherical > solid sphere > solid cylinder > hoop (c) Rank the objects' rotational kinetic energies from highest to lowest as the objects roll down the ramp. 1. solid cylinder > thin spherical > solid sphere > hoop 2. hoop > thin spherical > solid cylinder > solid sphere 3. hoop > solid cylinder > solid sphere > thin spherical 4. thin spherical > solid sphere > solid cylinder > hoop
If a solid and a hollow sphere of same mass m relesed from a smooth inclined plane,which will reach down first
The solid sphere will reach down first because it has a smaller rotational inertia. For the solid sphere, rotational inertia I = (2/5)*MR^2 for horrow I = (2/3)*MR^2. Where M is mass and R is radius. The kinetic energy is K = (1/2)*I*w^2, where w is the angular velocity, I is the inertia. K is the same for both sphere since they have the same mass and position on the plane and this gives the same potential energy. When the spheres roll, this potential energy is transfered to kinetic energy k. From the equation of K, a larger "I" gives a smaller "W", and 'W' is higher for smaller "I" as K is same. so the solid sphere must reach down first as it has a smaller rotational inertia and this cause the sphere to roll faster than the other as "w" is higher.
A solid cylinder of m mass rolls on a rough inclined plane with a β angle without slipping. What is the friction required for rolling?
Hey there! First of all such a good question! Had to use so many concepts to get here…Here goes,This is a diagram of the situation. Now, next will be a free body diagram of the Solid Cylinder.Now, by looking at the situation, we can say that the normal force will be balanced by Mgcosβ, while friction will apply in the opposite direction of the motion. We know that rolling is a motion of the body made possible due to torque, so let's calculate the torque due to all the forces here.(consider radius of the cylinder to be ‘r’ and point of rotation to be the Centre of the mass of the sphere)Torque due to Normal force and Mg (and all its components) will be zero (because they pass through the point of rotation). But, torque due to friction would be =(f*r) (force by perpendicular distance).Also, we know thatTorque = Moment of inertia*Angular accelerationThus we getτ=IαBut τ=f*r and I(for solid cylinder is) = 1/2 M*r² and also α=a/rThus, we get, after substituting the values for f, I, and a;μ(Mgcosβ) * r = 1/2 Mr² * gsinβ/rSolving which gives usμ=1/2 tanβCorrect me if wrong…Edit : Not able to upload the pics due to unstable Internet. Will try in the morning till then, imagine a slope with the sphere rolling down. I've used the forces Mg(gravitational) , normal reaction force and friction.. Made components of Mg and balanced them.Edit 2: uploaded the pics
Three objects of uniform density- a solid sphere, a solid cylinder, and a hollow cylinder- are placed at?
momen inertia of (each object is solid) I_sphere = (2/5)mr² I_cylinder = (1/2)mr² I_thin_cylinder = mr² torque = I α (friction) r = I (a/r) friction = Ia/r² second Newton's law of linear motion, ΣF = m a m g sin θ - friction = m a m g sin θ - Ia/r² = m a m g sin θ = (m + I/r²)a a = m g sin θ/(m + I/r²) a ∼ 1/I acceleration inversely proportional with respect to the moment of inertia, I_sphere < I_cylinder < I_thin_cylinder The sphere reaches the bottom for the first time and thin_cylinder for the last one. 'Try this at home and note that the result is independent of the masses and the radii of the objects.' moment of inertia I = kmr² hence k is a real constant. a = m g sin θ/(m + I/r²) a = m g sin θ/(m + kmr²/r²) a = g sin θ/(1 + k) acceleration a is independent of the masses and the radii of the objects.