Question

A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, without slipping, from the top of an inclined plane that is 1.9 m above the ground. Find the final linear velocity of the thin cylindrical shell.The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

Answer 1

Answer:

v_2,shell = 6.1056 m/s

Explanation:

Given:

- height of the top of inclined plane h = 1.9 m

- Mass Inertia (shell) I_shell = mR^2

Where R: radius and m: mass of the shell

Find:

Final linear velocity of shell v_2,shell

Using Energy balance:

                             E_p,1 + E_k,1 = E_p,2 + E_k,2

Given that E_k,1 = 0 and E_p,2 = 0 released from rest and reach ground respectively. Hence,

                                          E_p,1 = E_k,1

                                m*g*h = 0.5 * I_shell * w^2

Loss in potential leads to gain in rotational energy of shell. Note this case is a simplification of general motion where the objects slips + rolls. Hence, without slipping:

                                             w = v_2 / r

                                          g*h = v_2^2 / 2

                                         v_2 = sqrt (2*g*h)

                                   v_2 = sqrt ( 2 * 9.81 * 1.9)

                                      v_2 = 6.1056 m /s

Answer 2

The final linear velocity of the thin cylindrical shell is 4.32 m/s

Given that:

• The height of the inclined plane = 1.9 m

Let consider the rotational inertia of the cylinder shell;

i.e.

• I₁ = mR²

We are also told that there both the cylinders are rolled down without slipping.

∴

• v₁ = ω₁R

• ω₁ = v₁ /R

Taking the conservation of energy, we have:

$\mathbf{\dfrac{1}{2} I_1 \omega_1 ^2 + \dfrac{1}{2} m(\omega_1 R)^2 = mgH }$

$\mathbf{\implies \dfrac{1}{2} mR^2(\dfrac{v_1}{R})^2 + \dfrac{1}{2} mv_1^2 = mgH }$

By making the final linear velocity (v₁) the subject; we have:

$\mathbf{v_1 =\sqrt{gH}}$

$\mathbf{v_1 =\sqrt{9.8 m/s^2 \times 1.9 m}}$

$\mathbf{v_1 =4.32 m/s}$

Learn more about rotational inertia here:

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