Question

A uniform, solid cylinder of radius 5.00 cm and mass 3.00 kg starts from rest at the top of an inclined plane that is 2.00 m long and tilted at an angle of 25.0° with the horizontal and rolls without slipping down the ramp. What is the cylinder's speedat the bottom of the ramp?

Answer 1

To solve this problem we will apply the principle of conservation of energy, for which the initial potential and kinetic energy must be equal to the final one. The final kinetic energy will be transformed into rotational and translational energy, so the mathematical expression that approximates this deduction is

KE_i+PE_i = KE_{trans}+KE_{rot} +PE_f

$KE_i = 0$, since initially cylinder was at rest

$PE_f = 0$ since at the ground potential energy is zero

The mathematical values are,

$mgh = \frac{1}{2} mV^2 + \frac{1}{2}I\omega^2$

Here,

m = mass

g= Gravity

h = Height

V = Velocity

$I = \frac{mr^2}{2}$ moment of Inertia in terms of its mass and radius

$\omega = \frac{V}{r}$ Angular velocity in terms of tangential velocity and its radius

Replacing the values we have that

mgh = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{mr^2}{2})(\frac{v}{r})^2

gh = \frac{v^2}{2}+\frac{v^2}{4}

v = \sqrt{\frac{4gh}{3}}

From trigonometry the vertical height of inclined plane is the length of this plane for $sin\theta$, then

$h = 2.00*sin 25$

$h = 0.845 m$

Replacing,

$v = \sqrt{\frac{4(9.8)(0.845)}{3}}$

$V = 3.32 m/s$

Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s

Answer 2

"The speed where an object changes its motion," says Velocity, a vector quantity. In this, "3.32 m/sec" Velocity of the cylinder at bottom of the ramp.

Cylinder's speed Calculation:

First, at the top or bottom of the ramp, you can save energy by doing so.

KEi + PEi = KEtrans. + KErot + PEf

KEi = 0, because the cylinder was initially at a standstill

PEf = 0, Potential energy is zero at the ground.

So

m×g×h = 0.5×m×V² + 0.5×I ×w²

w = V/r

I = cylindrical moment of inertia = m× r²/ 2

m×g×h = 0.5×m×V² + 0.5×(m×r²/2)×(v/r)²

g×h = V²/2 + V²/4

V =√(4×g×h/3)

where = h = Angled plane vertical height = length of plane×sin theta (from trigonometry)

h = 2.00×sin 25° = 0.845 m

So,

V = √(4×9.81×0.845/3)

V = 3.32 m/sec = At the bottom of the ramp, the cylinder's velocity

Find out more about the cylinder here:

https://brainly.com/question/20895732?referrer=searchResults