Question

A 30-cm tall solid cylinder of diameter 20 cm is standing on the ground. If the force exerted on the ground by the weight of the cylinder is 500 N, what is the density of the material of the cylinder? A solid sphere of mass M and radius R (l=(2/5)MR^2) is placed on lop of a friction less inclined The plane is 7.0 m long and is making an angle of 35 degrees from the horizontal. Assuming the starts from rest at the top and rolls down the plane without sliding, find its linear velocity when it the bottom of the plane.

Physics help

Answer 1

11.) Total volume of the cylinder

$=\pi r^2h= \pi \times 0.1^2 \times 0.3m^3=9.425 \times 10^{-3}m^3$

If the force exerted on the ground by the weight of the cylinder is , then mass of the cylider is

$=\frac{500}{9.8}kg=51.02kg$

Therefore, density of the material of the cylinder

$\frac{51.02}{9.425 \times 10^{-3}}kg/m^3=5413.26kg/m^3=5.413gm/cc$

12.) Component of the force acting on the sphere perpendicular to the plane = $Mg\cos 35^0$

Component of the force acting on the sphere parallel to the plane = $Mg\sin 35^0$

Torque= Moment of inertia x angular acceleration

$N=F \cdot R=I \cdot \alpha \\ \\ \therefore \alpha=\frac{F\times R}{I}=\frac{Mg \sin 35^0 \times R}{\frac{2}{5}MR^2}=\frac{5}{2R}g \sin 35^0$

Linear acceleration of the sphere

$a=\alpha \cdot R=\frac{5}{2}g \sin 35^0$

Linear velocity of the sphere when it will reach the ground will be

$v^2=u^2+2as \downarrow \\ \\ v=\sqrt{2 \times a \times s}=\sqrt{2 \times \frac{5}{2}g \sin 35^0 \times 7} m/s^2=14.03ms^{-2}$