Question

A cylinder of mass M and radius r rolls down a ramp at height of 2m above the ground and eventually takes off the ramp 20cm above the ground at an angle of 40degrees from the x-axis. Assuming the cylinder didn't slipped as it rolls down, calculate the horizantal range at which the cylinder landed on the ground.

Answer 1

Answer on Question #63557, Physics / Mechanics | Relativity

A cylinder of mass M and radius r rolls down a ramp at height of 2m above the ground and eventually takes off the ramp 20cm above the ground at an angle of 40degrees from the x-axis. Assuming the cylinder didn't slipped as it rolls down, calculate the horizontal range at which the cylinder landed on the ground.

Solution:

Find the speed with which a cylinder rolling on ramp

$$m g h = \frac {m v ^ {2}}{2} + \frac {J \omega^ {2}}{2}$$ $$J = \frac {2}{5} m R ^ {2}$$ $$\omega = \frac {v}{R}$$ $$m g h = \frac {m v ^ {2}}{2} + \frac {v ^ {2}}{2 R ^ {2}} \cdot \frac {2 m R ^ {2}}{5}$$ $$m g h = \frac {7}{1 0} m v ^ {2}$$ $$v = \sqrt {\frac {1 0 g h}{7}}$$ $$v = \sqrt {\frac {1 0 \cdot 9 . 8 \cdot 2}{7}} = 5. 3 m / s$$$$v _ {x} = v \cos 4 0 {}^ {\circ} = 5. 3 \cdot 0. 7 7 = 4. 1 m / s$$ $$v _ {y} = v \sin 4 0 {}^ {\circ} 5. 3 \cdot 0. 6 4 = 3. 4 m / s$$ $$\Delta y = v _ {y} t + \frac {1}{2} g t ^ {2}$$ $$0. 2 = 3. 4 t + 4. 9 t ^ {2}$$ $$4. 9 t ^ {2} + 3. 4 t - 0. 2 = 0$$ $$t = 0. 0 5 5 s$$ $$\Delta x = v _ {x} t$$

Where $\Delta x = R$ , $\Delta x = 4.1 \cdot 0.055 = 0.23m$

Answer: 0.23 m

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