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what is the final velocity (in m/s) of a hoop that rolls without slipping down a 6.50-m-high hill, starting from rest?

User Dineshkani
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1 Answer

3 votes

Answer:

Approximately
7.99\; {\rm m\cdot s^(-1)}.

(Assuming that
g = 9.81\; {\rm N \cdot kg^(-1)} and that the thickness of the loop is negligible.)

Step-by-step explanation:

Let
m denote the mass of the hoop, and let
r denote its radius.

Under the assumptions, the moment of inertia of this hoop would be:


\displaystyle I = m\, r^(2).

Let
v denote the linear velocity of the hoop at the bottom of the hill. The linear kinetic energy of the hoop would be:


\displaystyle (1)/(2)\, m\, v^(2).

Since the hoop is rolling without slipping, its angular velocity would be
\omega = v / r. The rotational kinetic energy of the hoop would be:


\begin{aligned}(1)/(2)\, I\, \omega^(2) &= (1)/(2)\, (m\, r^(2))\, \left((v)/(r)\right)^(2) \\ &= (1)/(2)\, (m\, r^(2)\, v^(2))/(r^(2)) \\ &= (1)/(2)\, m\, v^(2)\end{aligned}.

The total kinetic energy of the hoop (linear and rotational) would be:


\begin{aligned}& (1)/(2)\, m\, v^(2) + (1)/(2)\, I\, \omega^(2) \\ =\; & (1)/(2)\, m\, v^(2) + (1)/(2)\, m\, v^(2) \\ =\; & m\, v^(2) \end{aligned}.

Assuming that total mechanical energy is conserved. Change in the Kinetic energy that the loop has gained would be the opposite of the change in the gravitational potential energy (GPE):


\begin{aligned}(\text{change in GPE}) &= m\, g\, \Delta h\end{aligned},

Where:


  • g = 9.81\; {\rm N\cdot kg^(-1)} by assumption, and

  • \Delta h = (-6.50)\; {\rm m} is the change in the height of the hoop.

By the conservation of energy:


(\text{change in KE}) + (\text{change in GPE}) = 0.


m\, v^(2) + m\, g\, \Delta h = 0.

Solve for
v:


\begin{aligned}m\, v^(2) &= m\, g\, (-\Delta h)\end{aligned}.


\begin{aligned}v &= √(g\, (-\Delta h)) \\ &= √((9.81)\, (-(-6.50)))\; {\rm m\cdot s^(-1)} \\ &\approx 7.99\; {\rm m\cdot s^(-1)}\end{aligned}.

In other words, the velocity of the loop would be approximately
7.99\; {\rm m\cdot s^(-1)} at the bottom of the hill.

User Djot
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