The velocity of cylinder which is released from rest on degree incline after it has rolled through a distance 1 is given by____________
KE= KE(transl) + KE (rot) = m•v²/2 + I•ω²/2
I =m•R²/2; ω=v/R.
KE= 3• m•v²/4
PE= m•g•h=m•g•s•sinα
PE=KE,
g•s•sinα=3•v²/4.
v=sqrt(4•g•s•sinα/3)
To determine the velocity of a cylinder that is released from rest on a degree incline and has rolled through a distance of 1, we can make use of the principles of rotational and translational motion.
Firstly, we need to understand the concept of rotational kinetic energy. When a cylinder is rolling without slipping, its rotational kinetic energy (KRot) is given by:
KRot = (1/2) * I * ω^2
Where I is the moment of inertia of the cylinder and ω is its angular velocity.
For a solid cylinder of mass m and radius r, the moment of inertia (I) is given by:
I = (1/2) * m * r^2
Next, we consider the translational kinetic energy (KTrans) of the cylinder. It is given by:
KTrans = (1/2) * m * v^2
Where v is the linear velocity of the cylinder.
In the absence of any external forces acting on the cylinder, the total mechanical energy (E) is conserved. Therefore, the sum of the rotational and translational kinetic energies is equal to the potential energy (mgh) of the cylinder at the starting position:
E = KRot + KTrans = mgh
Since the cylinder is released from rest, its initial velocity is zero. Thus, its initial kinetic energy is zero, and only the potential energy term remains:
E = mgh
Now, we can solve for the final velocity (v) of the cylinder. We need to know the height (h) of the incline, the mass (m) of the cylinder, and the acceleration due to gravity (g). Once these values are known, we can rearrange the equation to solve for v:
v = √(2gh)
Therefore, the velocity of the cylinder, which is released from rest on a degree incline after it has rolled through a distance of 1, is given by v = √(2gh), where h is the height of the incline and g is the acceleration due to gravity.