A solid uniform cylinder of mass 4.1 kg and radius 0.057 m rolls without slipping at a speed of 0.79 m/s. What is the cylinder’s total kinetic energy?

Question

asked Dec 6, 2020 140k views

1 Answer

Cmcquillan · Answer 1 · 2020-12-12T20:55:41+0000

Answer:

The cylinder’s total kinetic energy is 1.918 J.

Step-by-step explanation:

Given that,

Mass = 4.1 kg

Radius = 0.057 m

Speed = 0.79 m/s

We need to calculate the linear kinetic energy

Using formula of linear kinetic energy

K.E_(l)=(1)/(2)mv^2

K.E_(l)=(1)/(2)*4.1*(0.79)^2

$K.E_(l)=1.279\ J$

We need to calculate the rotational kinetic energy

$K.E_(r)=(1)/(2)* I\omega^2$

K.E_(r)=(1)/(2)*(1)/(2)* mr^2*((v)/(r))^2

K.E_(r)=(1)/(4)* m* v^2

K.E_(r)=(1)/(4)*4.1*(0.79)^2

$K.E_(r)=0.639\ J$

The total kinetic energy is given by

K.E=K.E_(l)+K.E_(r)

K.E=1.279+0.639

$K.E=1.918\ J$

Hence, The cylinder’s total kinetic energy is 1.918 J.