Answer:
![46620\ \text{N}](https://tex.z-dn.net/?f=46620%5C%20%5Ctext%7BN%7D)
Explanation:
m = Mass of roller coaster = 2000 kg
r = Radius of loop = 24 m
v = Velocity of roller coaster = 18 m/s
g = Acceleration due to gravity = ![9.81\ \text{m/s}^2](https://tex.z-dn.net/?f=9.81%5C%20%5Ctext%7Bm%2Fs%7D%5E2)
Normal force at the point will be
![N-mg=\dfrac{mv^2}{r}\\\Rightarrow N=\dfrac{mv^2}{r}+mg\\\Rightarrow N=\dfrac{2000\times 18^2}{24}+2000\times 9.81\\\Rightarrow N=46620\ \text{N}](https://tex.z-dn.net/?f=N-mg%3D%5Cdfrac%7Bmv%5E2%7D%7Br%7D%5C%5C%5CRightarrow%20N%3D%5Cdfrac%7Bmv%5E2%7D%7Br%7D%2Bmg%5C%5C%5CRightarrow%20N%3D%5Cdfrac%7B2000%5Ctimes%2018%5E2%7D%7B24%7D%2B2000%5Ctimes%209.81%5C%5C%5CRightarrow%20N%3D46620%5C%20%5Ctext%7BN%7D)
The force exerted on the track is
.
Answer:
![F=133N](https://tex.z-dn.net/?f=F%3D133N)
Explanation:
From the question we are told that:
Length ![l=3.0m](https://tex.z-dn.net/?f=l%3D3.0m)
Mass ![m=24kg](https://tex.z-dn.net/?f=m%3D24kg)
Distance from Tip ![d=35cm](https://tex.z-dn.net/?f=d%3D35cm)
Generally, the equation for Torque Balance is mathematically given by
![mg(l/2)=F(l-d)](https://tex.z-dn.net/?f=mg%28l%2F2%29%3DF%28l-d%29)
![2*9.81(3/2)=F(3-35*10^-2)](https://tex.z-dn.net/?f=2%2A9.81%283%2F2%29%3DF%283-35%2A10%5E-2%29)
Therefore
![F=133N](https://tex.z-dn.net/?f=F%3D133N)
Answer:
D
Explanation:
From the information given:
The angular speed for the block ![\omega = 50 \ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%2050%20%5C%20rad%2Fs)
Disk radius (r) = 0.2 m
The block Initial velocity is:
![v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s](https://tex.z-dn.net/?f=v%20%3D%20r%20%5Comega%20%5C%5C%20%5C%5C%20%20v%20%3D%20%280.2%20%20%5Ctimes%2050%29%20%5C%5C%20%5C%5C%20%20v%3D%2010%20%5C%20m%2Fs)
Change in the block's angular speed is:
![\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s](https://tex.z-dn.net/?f=%5CDelta%20_%7B%5Comega%7D%20%3D%20%5Comega%20-%200%20%5C%5C%20%5C%5C%20%3D%2050%20%5C%20rad%2Fs)
However, on the disk, moment of inertIa is:
![I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2](https://tex.z-dn.net/?f=I%3D%20mr%5E2%20%5C%5C%20%5C%5C%20I%20%3D%20%283%20%5Ctimes%200.2%5E2%29%20%5C%5C%20%5C%5C%20I%20%3D%200.12%20%5C%20kgm%5E2)
The time t = 10s
∴
Frictional torques by the wall on the disk is:
![T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m](https://tex.z-dn.net/?f=T%20%3D%20I%20%5Ctimes%20%28%5Cdfrac%7B%5CDelta_%7B%5Comega%7D%7D%7Bt%7D%29%20%5C%5C%20%5C%5C%20%3D%200.12%20%5Ctimes%20%28%5Cdfrac%7B50%7D%7B10%7D%29%20%20%5C%5C%20%5C%5C%20%3D0.6%20%5C%20N.m)
Finally, the frictional force is calculated as:
![F = \dfrac{T}r{}](https://tex.z-dn.net/?f=F%20%3D%20%5Cdfrac%7BT%7Dr%7B%7D)
![F= \dfrac{0.6}{0.2} \\ \\ F = 3N](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B0.6%7D%7B0.2%7D%20%5C%5C%20%5C%5C%20F%20%3D%203N)