1.A 3.9 kg stone is tied to a 5.8 m long string and swung around a circle at a c

Question

1.A 3.9 kg stone is tied to a 5.8 m long string and swung around a circle at a constant angular velocity 1.7 1/s. Find the magnitude of the torque about the origin

2.A uniform meter stick is placed on and parallel to the x axis, with the 50 cm mark at the origin. It is pivoted at the 50 cm mark. A 3.9 N force is applied at the 43 cm mark, and a -4.1 N force is applied at the 90 cm mark. Find the net torque on the rod.

3.
A thin disk, mounted on a frictionless vertical shaft of negligible rotational inertia, is rotating at 215 revolutions per minute. An identical disk (that is not initially rotating) is dropped onto the first disk. The frictional force between the disks causes them to rotate at a common angular velocity. Find this common angular velocity.
Use rad/s for your unit.

4.
A merry-go-round has a radius of 10 m and a moment of inertia of 663 kg.m2. It is initially spinning with an angular velocity ? = 1.4 rad/s when a 18 kg child crawls from the center to the rim. When the child reaches the rim, what is the angular velocity of the merry-go-round?
Use rad/s for your answer.

5.At the same   instant that a 2.7 kg ball is dropped from 30 m above the Earth's   surface, a second ball, with a mass of 8.8 kg, is thrown straight   upward from the Earth's surface with initial velocity of 1.6 m/s. They   move along nearby lines and pass without colliding. At the end of 3   s, what is the height above the Earth of the center of mass of the two-ball system?

6.
At the same   instant that a motionless 9.6 kg ball is dropped from 27 m above the Earth's   surface, a second ball, with a mass of 4.5 kg, is thrown straight   upward from the Earth's surface with initial velocity of 21.1 m/s. They   move along nearby lines and pass without colliding. At the end of 2.1   s, what is the velocity of the center of mass of the two-ball system?

Choose down as the -y direction.

7.
At the same instant that a 6.2 kg ball is dropped from 11 m above the Earth's surface, a second ball, with a mass of 2.2 kg, is thrown straight upward from the Earth's surface with initial velocity of 19 m/s. They move along nearby lines and pass without colliding. At the end of 2.3 s, what is the accleration of the center of mass of the two-ball system?

8.
Block A, with a mass of 2.4 kg, is stationary, while block B, with a mass of 8.6 kg, is moving at 8.1 m/s. What is the speed of the center of mass of this two-block system?

9.
A 932 kg rocket is motionless in space. To start  moving, its main engine is fired for 7 s, during which time it ejects exhaust gases at 1205 m/s. At the end of this process it is moving at 17 m/s. What is the approximate mass of ejected gas?

10.
The center of mass of a system consisting of two spheres is moving in the +x direction with a speed of 6.6 m/s. One sphere has a mass of 9.8 kg and the other has a mass of 7.8 kg. What is the total momentum of the system?

Explanation / Answer

1)

the Torque about the origin ia zero, since the angular velocity is constant.T = Ia a =0 => T=0

2)

T = 3.9*7 -(-4.1*40) = 1.913 Nm

3)

angular momentum is conserved. initial Ang momentum = IW1

final = 2IW2. equating IW1 = 2IW2 => W2 = W1/2 = 215/2 = 107.5 revolutions/s

4)

angular momentum is conserved. initial angular momentum = IW1

final = IW2 + I2W2

663*1.4 = 663*W2 + 18*10^2*W2

W2 = 663*1.4/2463 = 0.376 rad/s

5)

initial height pf the centre of mass of the system = 2.7*30+8.8*0/(2.7+8.8) = 7.04 m

acceleration of centre of mass = g = 10 m/s^2

height after 3 seconds = 7.04 - 0.5gt^2 = 7- 45 = -38 m

since the height cant be negative, it means that both balls hit the ground, and cnetre of mass is at zero height from the ground

6)

initial velocity pf the centre of mass of the system = (9.6*0 + 4.5*21.1)/(4.5+9.6) = 6.73 m/s upwards

aacceleration of 2 ball system = g = 10 m/s^2

at the end of 2.1 seconds, velocity = 6.73 - gt = 6.73 - 21 = -4.27 m/s

velocity is 4.27 m/s downwards

7)

acceleration of the balls that are in air is g - 10m/s^2

at the end of 2.3 s, the 6.2 kg ball dropped from 11m would have hit the ground since 11 < 0.5*10*2,3^2. so its acceleration is zero

the 2.2 kg ball would still be in the air since 2.3 < 2*19/10= 3.8 s

so the acceleration of 2.2 kg bAll = 10 m/s^2

acceleration of centre of mass = (6.2*0+2.2*10)/(2.2+6.2) = 2.61 m/s^2

8)

speed = (M1V1+M2V2)/(M1+M2) = (2.4*0 + 8.6*8.1)/(2.4+8.6) = 6.33 m/s

9)

Lets say the mass of the ejected gas is x.

the final mas of the rocket = 932 -x kg

final speed = 17 m/s

final momentum = (932 - x)*17 kgm/s

this momentum must equal the momentum of the ezhaust gases released

= x*1205 kgm/s

equating (932 - x)*17 = x*1205 => x = 932*17/1222 = 13 kg

mass of exhaust gases = 13kg

10)

total momentum = total mass* speed of centre of mass

= (7.8+9.8)* 6.6 = 116.16 kgm/s

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