A hollow spherical shell 2 kg mass 35 cm radius I 23mr2 is
A hollow spherical shell [2 kg mass, 35 cm radius, I = (2/3)mr^2] is released from rest. It rolls without slipping down a slope [A = 3.5 m], through a circular are [80 cm radius, beta = 57 degrees, alpha = 30 degrees - both from the vertical] and along a slope [B = 5 m] before becoming airborne and landing on the ground. H = 0.75 m. Set the ground to be 0 J. Calculate the magnitude of the normal force acting on the shell at the lowest point of the circular arc. Calculate the number of rotations made by the shell as it travels from the point of release to the top of slope B. Calculate the shell\'s angular speed at the apex of flight and the height of the apex relative to the ground. Calculate the shell\'s range and kinetic energy at the moment of impact.
Solution
1) Fnet = mg +mV2/r
You need to calculate the height , and get the velocity by PE=KE using the linear velocity + rotational velcotiy KE
2) no of rotations, without slipping = total length divided by circumf of the sphere
3) at take off from the guide it starts a parabolic flight. Use projectile
4)same. This is part of a quiz given by Prof Mona of SD mesa. You really need to do this on our own.
![A hollow spherical shell [2 kg mass, 35 cm radius, I = (2/3)mr^2] is released from rest. It rolls without slipping down a slope [A = 3.5 m], through a circular A hollow spherical shell [2 kg mass, 35 cm radius, I = (2/3)mr^2] is released from rest. It rolls without slipping down a slope [A = 3.5 m], through a circular](/WebImages/1/a-hollow-spherical-shell-2-kg-mass-35-cm-radius-i-23mr2-is-965443-1761494824-0.webp)