Use g = 10m/s 2 1.Two masses are attached to a light rod, i.e. neglect the mass

Question

Use g = 10m/s2

1.Two masses are attached to a light rod, i.e. neglect the mass of the

rod, and they rotate about a vertical axis through the center of mass

with an angular velocity of 6rad/s. m1=2kg and is 0.5m from the vertical axis while m2=1kg and is 1m from the vertical axis.

a)What is the moment of inertia about that axis?

Â

b)What is the kinetic energy of the system?

Â

c)What is the angular momentum of the system?

Â

2. A disk is initially rotating with an angular velocity of 2rad/s. Over the next 5s it rotates

through three complete revolutions. What was its angular acceleration for those 5s?

Â

3. The weight of a block in air is 20N. The apparent weight when it is submerged in oil (Ã=800kg/m3) is 15N

Â

a)What is the buoyant force on the block when it is submerged?

Â

b)What is the volume of oil displaced when the block is submerged?

Â

c)What is the density of the block?

Â

4. A hollow sphere of radius R and mass M is rolling without slipping across the floor with a linear velocity of v1. It encounters an incline that rises up a vertical

distance 2m. At the top of the rise its velocity is v2=3m/s.

What was v1?

Â

5. A pole of mass 15kg and length 3m is attached to a vertical wall

by a hinge that is free to rotate. It makes and angle of 70o

with the wall. A wire is attached to the end of the pole and the wire makes

an angle of 50o with the wall. A sign of mass 10kg hangs from

the end of the pole. What must the tension in the wire be to

support the pole and sign?

Explanation / Answer

1.

A)m1=2kg d1 = 0.5m , m2=1kg d2= 1m

I total = m1d1^2 + m2d2^2 = 1.5 kg m^2

B) w = 6 rad/s

Kinetic energy = 0.5 I w^2 = 0.5 * 1.5 * 36 = 27 J

C) ANgular momentum = Iw = 1.5 * w = 9

2)

intial w = 2 rad/s

For next 5 sec it takes 3 rev => 6 * pi radians

6 pi = 2 * 5 +0.5 acc. 5 ^ 2 = 10 + 12.5 acc = 6 * pi =18.846

Acceleration = 0.707 rad/s2

3) a)Bouyant force = real weight - apparent = 20 - 15 = 5 N

b) Volume = weight of displaced fluid / density

weight of displaced fluid = Bouyant force

Volume = (5/10)/800 = 0.625 * 10^-3 m3

Density/density of fluid = weight / weight of displaced fluid = 20/5 = 4

Density = 800 * 4 = 3200 kg/m3

4) Moment of inertia at the point of contact = 5/3 MR^2

gain in potetntial energy = loss in kinetic enrgy

M * 10 * 2 = 0.5 I ((v1/R)^2 - (3/R)^2)

20 = 5/6 *( v1^2 - 9)

v1 = sq.rt(33)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.