A small 0.300 kg bird is flying horizontally at 3.30 m/s toward a 0.750 kg thin
ID: 1465996 • Letter: A
Question
A small 0.300 kg bird is flying horizontally at 3.30 m/s toward a 0.750 kg thin bar hanging vertically from a hook at its upper end, as shown in the figure below (Figure 1) .
Part A
When the bird is far from the bar, what are the magnitude and direction (clockwise or counterclockwise) of its angular momentum about a horizontal axis perpendicular to the plane of the figure and passing through point A?
Express your answer in kilogram meters squared per second to three significant figures. Let counterclockwise angular momentum be positive, clockwise angular momentum be negative.
SubmitMy AnswersGive Up
Part B
When the bird is far from the bar, what are the magnitude and direction (clockwise or counterclockwise) of its angular momentum about a horizontal axis perpendicular to the plane of the figure and passing through point B?
Express your answer in kilogram meters squared per second to three significant figures. Let counterclockwise angular momentum be positive, clockwise angular momentum be negative.
SubmitMy AnswersGive Up
Part C
When the bird is far from the bar, what are the magnitude and direction (clockwise or counterclockwise) of its angular momentum about a horizontal axis perpendicular to the plane of the figure and passing through point C?
Express your answer in kilogram meters squared per second to three significant figures. Let counterclockwise angular momentum be positive, clockwise angular momentum be negative.
SubmitMy AnswersGive Up
Part D
When the bird is just ready to hit the bar, but is still flying horizontally, what are the magnitude and direction (clockwise or counterclockwise) of its angular momentum about a horizontal axis perpendicular to the plane of the figure and passing through point A?
Express your answer in kilogram meters squared per second to three significant figures. Let counterclockwise angular momentum be positive, clockwise angular momentum be negative.
SubmitMy AnswersGive Up
Part E
When the bird is just ready to hit the bar, but is still flying horizontally, what are the magnitude and direction (clockwise or counterclockwise) of its angular momentum about a horizontal axis perpendicular to the plane of the figure and passing through point B?
Express your answer in kilogram meters squared per second to three significant figures. Let counterclockwise angular momentum be positive, clockwise angular momentum be negative.
Part F
When the bird is just ready to hit the bar, but is still flying horizontally, what are the magnitude and direction (clockwise or counterclockwise) of its angular momentum about a horizontal axis perpendicular to the plane of the figure and passing through point C?
Express your answer in kilogram meters squared per second to three significant figures. Let counterclockwise angular momentum be positive, clockwise angular momentum be negative.
L = kg?m2/s L = kg?m2/s L = kg?m2/s L = kg?m2/s L = kg?m2/s L =Explanation / Answer
mass= 0.300 kg
velocity,v= 3.30 m/s
mass of the wood=0.750 kg
The angular momentum is the linear momentum times the normal (perpendicular) distance to the line of motion:
Linear momentom is mass times velocity
Linear momentum =mass*velocity= 0.300*3.30=0.99
the distance to A = 1.50 m;
angular momentum = Linear momentom *distance=0.99*1.5=1.485 kg*m2/s; the direction is counterclockwise
B) The linear momentum is the same, but the normal distance to point B is zero, so the angular momentum is zero
C) The linear momentum is the same, but the distance to C is 1.80 - 1.50 = 0.30 m;
so the angular momentum =0.99*0.3= 0.297 kg*m2/s; The direction is clockwise
Since they are moving and not hit the bar. I think its is similiar to the cases that we discuss above
D)Linear momentum =mass*velocity= 0.300*3.30=0.99
the distance to A = 1.50 m;
angular momentum = Linear momentom *distance=0.99*1.5=1.485 kg*m2/s; the direction is counterclockwise
E) The linear momentum is the same, but the normal distance to point B is zero, so the angular momentum is zero
F) The linear momentum is the same, but the distance to C is 1.80 - 1.50 = 0.30 m;
so the angular momentum =0.99*0.3= 0.297 kg*m2/s; The direction is clockwise