When an object is turning a corner what direction is the acceleration?

1 answer:

7 0

If the corner is rounded and is perfectly circular, then the acceleration is centripetal and is always directed toward the center.

You might be interested in

Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what i

Arte-miy333 [17]

I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
$R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i$
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
$R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_x=0.96m$
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
$R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_y=-0.84m$
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
$v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}$
$v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}$
Total velocity is:
$v=\sqrt{v_x^2+v_y^2}$
Once we calculate velocities we get:
$v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}$
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing