Answer:
a)We know that acceleration a=dv/dt
So dv/dt=kt^2
dv=kt^2dt
Integrating we get
v(t)=kt^3/3+C
Puttin t=0
-8=C
Putting t=2
8=8k/3-8
k=48/8
k=6
Answer:
a. Rotational speed of the drill = 375.96 rev/min
b. Feed rate = 75 mm/min
c. Approach allowance = 3.815 mm
d. Cutting time = 0.67 minutes
e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min
Explanation:
Here we have
a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min
b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min
c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm
d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes
e. R = 0.25πD²fr = 9525 mm³/min.
Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember
To solve this problem we will apply the concepts related to translational torque, angular torque and the kinematic equations of angular movement with which we will find the angular displacement of the system.
Translational torque can be defined as,
![\tau = Fd](https://tex.z-dn.net/?f=%5Ctau%20%3D%20Fd)
Here,
F = Force
d = Distance which the force is applied
![\tau = (1N)(1m)](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%281N%29%281m%29)
![\tau = 1N\cdot m](https://tex.z-dn.net/?f=%5Ctau%20%3D%201N%5Ccdot%20m)
At the same time the angular torque is defined as the product between the moment of inertia and the angular acceleration, so using the previous value of the found torque, and with the moment of inertia given by the statement, we would have that the angular acceleration is
![\tau = I\alpha](https://tex.z-dn.net/?f=%5Ctau%20%3D%20I%5Calpha)
![\alpha = \frac{\tau}{I}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Ctau%7D%7BI%7D)
![\alpha = \frac{1N\cdot m}{100kg\cdot m^2}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1N%5Ccdot%20m%7D%7B100kg%5Ccdot%20m%5E2%7D)
![\alpha = 0.01rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.01rad%2Fs%5E2)
Now the angular displacement is
![\theta = \omega_0 t + \frac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega_0%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
Here
= Initial angular velocity
t = time
Angular acceleration
= Angular displacement
Time is given as 1 minute, in seconds will be
![t = 1m = 60s](https://tex.z-dn.net/?f=t%20%3D%201m%20%3D%2060s)
There is not initial angular velocity, then
![\theta= \frac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%20%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
Replacing,
![\theta= \frac{1}{2}(0.01)(60)^2](https://tex.z-dn.net/?f=%5Ctheta%3D%20%5Cfrac%7B1%7D%7B2%7D%280.01%29%2860%29%5E2)
![\theta = 18rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2018rad)
The question neglects the effect of gravitational force.
Yes it does have a large spot