Answer:
Explanation:
average speed more than 25.0m/s.
Answer:
20 cm to the right of the center or 20+50 = 70 cm from the left side.
Explanation:
The length of meter stick is 1 m = 100 cm
Balance point on 50 cm
From the center the 20 N weight is 50-20 = 30 cm
Torque is obtained when force is multiplied with the distance
As the force is conserved we have
![\dfrac{20}{30}\times 30=20\ cm](https://tex.z-dn.net/?f=%5Cdfrac%7B20%7D%7B30%7D%5Ctimes%2030%3D20%5C%20cm)
The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.
Answer:
0.231 N
Explanation:
To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be
![\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Ctheta%7D%7Bt%7D%20%3D%20%5Cfrac%7B6.37%7D%7B9.87%7D%20%3D%200.6454%20rad%2Fs%5E2)
If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:
![I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2](https://tex.z-dn.net/?f=%20I%20%3D%20%5Cfrac%7BmL%5E2%7D%7B3%7D%20%3D%20%5Cfrac%7B1.27%2A0.847%5E2%7D%7B3%7D%20%3D%200.303kgm%5E2)
According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is
![T = I\alpha = 0.303*0.6454 = 0.196 Nm](https://tex.z-dn.net/?f=%20T%20%3D%20I%5Calpha%20%3D%200.303%2A0.6454%20%3D%200.196%20Nm)
So the force acting on the other end to generate this torque mush be:
![F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N](https://tex.z-dn.net/?f=%20F%20%3D%20%5Cfrac%7BT%7D%7BL%7D%20%3D%20%5Cfrac%7B0.196%7D%7B0.847%7D%20%3D%200.231%20N)