The figure below gives the speed $v$ versus time $t$ for a $0.520$ kg object of radius $6.20$ cm that rolls smoothly down a $30^\circ$ ramp. The scale on the velocity axis is set by $v_s = 4.0\text{ m/s}$. What is the rotational inertia of the object?
Someone said the formula is, $$mgh = \frac{1}{2} mv^2 + \frac{1}{2} Iw^2$$
I don't see height in this problem, though. Can it be found?
Integrate the velocity to recover the distance. $$ \int v dt = x $$ Read the chart to recover the velocity function. The velocity is constant, and after $1$ second, is $3.5$ $m$/$s$: $$ v(t) = 3.5t $$ The distance traveled, in meters, is $$ \int_{0}^{1} 3.5t\, dt = \frac{7}{4}t^{2} \Big\lvert_{0}^{1} = \frac{7}{4}(1) - 0 = \frac{7}{4} $$
(Thanks to @user435909 for the proof read.)
Given:
$mgx = (1/2)mv^2 + (1/2)I \omega ^2$,
find $I$.
An object rolls down a ramp of 30° inclination. Let O be the starting point.
Consider an axis vertically downward through O, let it be the positive x-axis. Measure x along this axis.
Initially $t= 0, x(t=0) = 0$ and $v(t=0) = 0$, where $v(t)$ is the velocity of the particle moving down the ramp.
Energy conservation:
Loss of potential energy: $mgx$, counted positively vertically downward,
$=$
gain in kinetic energy $(1/2) mv^2 + (1/2) I\omega ^2$.
The kinetic energy consists of two parts:
Linear motion down the ramp, $(1/2) mv^2$, and rotational motion around the CM, $(1/2) I\omega^2$.
Let $s$ be the distance from $O$ down the ramp.
$Sin (30°) = 1/2 = x/s$, i.e. $s = 2x$.
We have:
(×) $mg(s/2) = (1/2)mv^2 + (1/2)I \omega^2$.
Let's look at the $v$-$t$ graph given in the problem.
We find $dv/dt = a$ (constant).
With the initial conditions $t=0, x(t=0) = 0$, and $v(t=0) =0$, integration yields:
1) $v(t) = at$;
2) $s(t) = (1/2) at^2$, and
eliminating $t$ to find $s =s(v)$:
3) $s(v) = (1/2)a(v(t)/a)^2 =(1/2) (v(t))^2 /a$.
Replacing $s$ in equation (×):
$(1/2)mgv^2/(2a) = (1/2) mv^2 + (1/2)I\omega^2$ .
With $\omega = r v$, the above equation becomes:
$mgv^2/(2a) = mv^2 + I r^2 v^2$.
Canceling $v$ and solving for $I$:
$$I = \frac{mg - 2am}{2ar^2}$$.
Check:
For $I = 0$ there is no rotational energy.
$a = g/2 = 5 [m/sec^2]$ which is the value of the $g$ component along the ramp, $a = g$ $cos (60°)$ .
In this problem $a = 3.5 [m/sec^2]$, less than $5 [m/sec^2]$ due to the rotation.
