Question

A man with a motorcycle driving on a straight road with a speed of 90 km / h . The mass of men along the bike of 100 kg .The officers chasing criminals from behind men using a helicopter , flying at an altitude of 500 m above the ground and at a speed of 144 km / h . The officer dropped the body of mass 400 kg ( without giving the initial speed of the helicopter ) . It turns out objects just overwrite the officer released the man .

In a system that consists of men along with the bike and the object is released attendant , find the acceleration of the center of mass and center of mass equation of motion

Answer 1

Answer to the question #57470, Physics / Mechanics | Relativity

A man with a motorcycle driving on a straight road with a speed of $90\mathrm{km} / \mathrm{h}$. The mass of men along the bike of $100\mathrm{kg}$. The officers chasing criminals from behind men using a helicopter, flying at an altitude of $500\mathrm{m}$ above the ground and at a speed of $144\mathrm{km} / \mathrm{h}$. The officer dropped the body of mass $400\mathrm{kg}$ (without giving the initial speed of the helicopter). It turns out objects just overwrite the officer released the man.

In a system that consists of men along with the bike and the object is released attendant, find the acceleration of the center of mass and center of mass equation of motion.

Answer

$$\mathrm{m}_{\mathrm{mot}} + \mathrm{m}_{\mathrm{men}} = 100\ \mathrm{kg},\ \mathrm{m}_{\mathrm{o}} = 400\ \mathrm{kg},\ \mathrm{v}_{\mathrm{mot}} = 90\ \mathrm{km} / \mathrm{h},\ \mathrm{v}_{\mathrm{hel}} = 144\ \mathrm{km} / \mathrm{h},\ \mathrm{v}_{\mathrm{hel}} = \mathrm{v}_{\mathrm{o}} \quad \text{— initial}$$

speed of the objects equals to the speed of helicopter

$$\overline{r}_{c} = \frac{(m_{\mathrm{mot}} + m_{\mathrm{men}})\overline{r}_{\mathrm{mot} + \mathrm{men}} + m_{\mathrm{o}}\ \overline{r}_{\mathrm{o}}}{m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{\mathrm{o}}} \quad \text{— position of the center of mass.}$$ $$\overline{v}_{c} = \frac{(m_{\mathrm{mot}} + m_{\mathrm{men}})\overline{v}_{\mathrm{mot}} + m_{\mathrm{o}}\ \overline{v}_{\mathrm{o}}}{m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{\mathrm{o}}} \quad \text{— speed of the center of mass.}$$ $$\begin{array}{l} x\text{-axis projection: } v_{c} = \frac{(m_{\mathrm{mot}} + m_{\mathrm{men}})v_{\mathrm{mot}} + m_{\mathrm{o}}v_{\mathrm{hel}}}{m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{\mathrm{o}}} \\ y\text{-axis projection: } v_{c} = \frac{-g t m_{o}}{m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{o}} \\ \overline{a}_{c} = \frac{(m_{\mathrm{mot}} + m_{\mathrm{men}})a_{\mathrm{mot}} + m_{\mathrm{o}}a_{\mathrm{o}}}{m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{\mathrm{o}}} \quad \text{— acceleration of the center of mass.} \\ x\text{-axis projection: } a_{c} = 0 \\ y\text{-axis projection: } a_{c} = \frac{-g m_{o}}{m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{o}} \\ \sum_{i} F_{i} = (m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{o})a_{c} = (m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{o})\frac{-g m_{o}}{m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{o}} = -g m_{o} \\ \end{array}$$ $$\sum_{i} F_{i} = -g m_{o} \quad \text{— center of mass equation of motion, where } \sum_{i} F_{i} \quad \text{— the sum of external forces.}$$ $$a_{c} = \frac{-g m_{o}}{m_{\mathrm{mot}} + m_{\mathrm{men}} + m_{o}} = -7.84\ \mathrm{m/s^{2}}$$

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