Question

A uniform cylinder of mass 4.3 kg and radius 0.4 m rolls down a ramp inclined at an angle 0.15 radians to the horizontal. what is the acceleration of the cylinder in m/s2 ?

Answer 1

The cylinder's acceleration is

$a=\frac{gsin}{1+\frac{I}{mrL} }$

$= 2g\frac{sin}{3}$θ

$= \frac{2(9.8) sin (0.15 rad)}{3} \\=0.9763 \frac{m}{sec r}$

How do you determine a cylinder's acceleration?

The cylinder's complete radius, $R$, from the center marks the contact point, therefore the torque created by friction is given by $fR = I\alpha$, where is the rotating acceleration. This rotational acceleration's corresponding linear acceleration, $\alpha$, is equal to $R$. The cylinder, which has its mass concentrated in its center, triumphs in the competition, followed by the disc and the hoop, with their respective final velocities being roughly $1.4:1.2:1$. We can also see that our findings are unaffected by the cylinders' masses.

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