A high 1 circular motion problem, a high 1 circular motion physics problem

Draw a circle of r, draw a line from the bottom to the top h, and the focus of the garden is the position of the small ring.

Direct column equation too.

mg/mn^2rcosa=tana

r-h/r=sina

For a Angle Synthesis.

g r-h) 1 2--- is the answer.

Did you make a mistake in the question?

According to v=r, and the linear velocity of each point on the edge of the two wheels is equal, so 2 n2 r2 60 = 2 n1 r1 60, i.e., n2 = n1 r1 r2

When R2=3cm, R1=, the speed of the driven wheel 2 is the smallest, and Nmin= X36=6R min

When r2=, the speed of the driven wheel 2 is the largest, and nmax= x36=216r min, so the speed change range of the driven wheel 2 is 6r min 216r min.

by v=r1 2 n1

r1=, vmin=

When r1=3cm, vmax=3x10 -2x2 x36 60=.

So the range of speed variation of the tape movement is.

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Angular velocity = 2pi period.

The cycle of the hand, minute hand, and second hand is 12 hours (12 x 60 minutes) to hold the letter, 1 hour (60 minutes), and 1 minute respectively.

Angular velocity 1 (12*60): 1 60:1=1:

The ratio of the speed of the 12:720 line segment is va:vb 1:

1The ratio of angular velocity is a: b=va r1:vb r2=r2:

The ratio of the R1 period is ta:tb and 2pi a:2pi b= b:

a=r1:r2

The ratio of rotational speed is Na:Nb=1 Ta:1 TB=R2:R1

O2 two wheels through friction transmission type chaotic rock, transmission between the two wheels do not slip, the ratio of the radius of the two wheels is R1: accompany pants R2, A, B are O1, O2 two wheels on the edge of the point, then A, B two points of the linear speed of the cycle is 12 hours (12 * 60 minutes), 1 hour (60 minutes) Bu Yu, 1 minute.

Angular velocity 1 (12*60): 1 60:1=1:12

It can be solved by the law of inertia, because the density of water is greater than that of the wax block, so the inertia of the water is greater than that of the wax block, when the test tube swings left and right, due to the greater inertia of the water, it will rush to the bottom of the test tube, so that the wax block will be pushed to the mouth of the test tube, so the wax block will move to A, select C.

They are all thrown outward, but the density of water is relatively large, and the centrifugal force received per unit volume is relatively large, so it is easier for water to be thrown out, so the wax block is squeezed in.

The object moving in the most circular motion needs a centripetal force, and a large unit volume mass needs a large centripetal force to do centrifugal motion. The small mass per unit volume requires a small centripetal force to do centripetal motion, (centrifuge principle).

Correct answer c

Because it shakes from side to side, the wax block is subjected to a downward force, and the decomposition force has a downward force, so it accelerates toward A.

1 The rotation period of the earth is (24*3600)=86400s; The radius of the Earth is 6400km, and the linear velocity of an object placed on the equator moving with the rotation of the Earth is (2* m s

2 The length of the minute hand of the watch is the distance from the axis to the tip of the minute hand of the watch, and the size of the linear velocity of the tip of the minute hand (2*

3 If the grinding wheel with a radius of 10cm and the rotation speed is 5r min, what is the angular velocity of the grinding wheel rotation (5*2) 60? Is the angular velocity equal at the points on the grinding wheel at different distances from the shaft? Is the linear velocity equal and unequal?

What is the linear velocity of the point on the edge of the grinding wheel?

4 The particle moves in a uniform circular motion around the fixed point of the state grip with radius r=, and the rotation speed n=3000r min, then the angular velocity of the particle is (3000*2) 60=314rad s, and the linear velocity is 314*

1. When the test tube is moving in a uniform circular motion, the ball is always at the bottom of the test tube, and the pressure of the ball on the test tube at the highest point is f:

f+mg=mrw^2

The pressure of the pellet on the tube at its lowest point is 3F:

3f-mg=mrw^2

Solution: f=mg

Either Eq. 1 or Eq. 2 yields the value of w.

w=20rad/s

2. W=30rad S:

At the highest point: F1+MG=MRW2

f1=35n

Nadir: F2-MG=MRW2

f2=55n

First of all, you need to figure out when the pressure on the bottom of the tube is the maximum value and when the pressure on the ball is the minimum value!

According to the analysis, the lowest point should be taken to the minimum value, n = mg-mrw 2, and at the lowest point n there is a maximum value, mg + mrw 2).

Now it's time for the valued!

If you don't understand why you get the maximum value at the highest point or the lowest, it's because the force is like that, and if it is in other cases, then the force has to be orthogonally decomposed).

Hope it helps.

The analysis can be leaked, only A rope is stressed, and W is the smallest; Only the B rope is stressed, and W is the largest.

The combined force of gravity and the large tensile force of the collapsed pure rope provides the centripetal force.

1. Only a is forced.

3／3mg＝mrw^2

r=1 (this can be seen from the empty shirt of the figure).

w=2, only b is forced.

mg＝mrw^2

w＝3．16rad/s

In summary: (rad s).

No, you understand the two different situations.

If you look at the diagram on the left below, it is the situation of placing the object on an inclined plane, at this time the object may be stationary or moving downward, and the direction of the vertical inclined plane is balanced by the force, and the supporting force n=mgcos

Looking at the diagram on the right, the resultant force of the two forces of the object supported by gravity and the inclined plane (ignoring the frictional force for the sake of convenience) is the centripetal force of circular motion for the object in the horizontal direction, at this time n=mg cos, and the centripetal force f=mg tan.

In this case, the vertical bevel direction is not balanced by force.

Compare the two cases.