Rotation Quiz
A 2kg block starts from rest on the positive x axis 3m from the origin and thereafter has an acceleration given by a=4i-3j in m/s^2. The torque, relative to the origin, acting on it at the end of 2s is:
-18N*m
A force F= 4.2i+3.7j+1.2k acts on a particle located at x=3.3m. What is the torque on the particle around the origin?
-Find cross product of 2 vectors: -4j+12k
A 2kg stone is tied to a 0.5m long string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75m from the origin. The magnitude of the torque about the origin is:
-Firstly it is required to find the centripetal force acting on the stone. It is given by : F=mrw^2 F=144N T=r*F=0.75*144 =108N*m
A playground merry-go-round has a radius of 3m and a rotational inertia of 600kg*m^2. It is initially spinning at 0.8 rad/s when a 20kg child crawls from the center to the rim. When the child reaches the rim the angular velocity of the marry-go-round is:
-conserve angular momentum = 0.62 rad/
A wheel starts from rest and has an angular acceleration that is given by a(t)=(6rad/s^4)t^2. After it as turned through 10 rev, its angular velocity is:
-integrate function to find velocity function with respect to time -integrate again to find position function -set position function to equal 20π -find t -plug into velocity function to find angular velocity
a single force acts on a particle P. Rank each of the orientations of the force shown below according to the magnitude of the time rate of change of the particle's angular momentum about the point O, least to greatest
1 and 2 tie, then 4, then 3
Three balls, with masses of 3M, 2M, and M, are fastened to a massless rod of length L as shown. The rotational inertia about the left end of the rod is:
3ML^2/2
Consider four objects, each having the same mass and the same radius: 1. a solid sphere 2. a hollow sphere 3. a flat disk in the x, y plane 4. a hoop in the x, y plane The order of increasing rotational inertia about an axis through the com and parallel to the z axis is:
4, 2, 3, 1 See rotational inertia formulas for diff shapes
Three identical balls are tied by light strings to the same rod and rotate around it, as shown below. Rank the balls according to their rotational inertia, least to greatest
A. 1, 2, 3 I=mr^2
Two identical disks, with rotational inertia I=1/2mr^2, roll w/o slipping across a horizontal floor and then up inclines. Disk A rolls up its incline w/o sliding. On the other hand, disk B rolls up a frictionless incline. Otherwise, the inclines are identical. Disk A reaches a height 12 cm above the floor before rolling down again. Disk B reaches a height above the floor of:
Answer: Disc B will reach to height of 8 cm height above the floor Explanation: As we know that two disc are identical and rolling on the horizontal surface Now we know that while disc is in rolling motion its kinetic energy is sum of rotational kinetic energy and transnational kinetic energy. So we have KE=1/2mv^2 + 1/2 I w^2 now we know that for pure rolling of disc KE=1/2mv^2+1/4mv^2 now when disc roll over the inclined surface then its total kinetic energy will convert into gravitational potential energy so we will have mv^2= 1/3mgH here we know that H = 12 cm now when another disc rolls up on frictionless inclined plane then it will lose all its translational kinetic energy but rotational kinetic energy will remain as it is as there is no torque on the disc So we have mgh=1/2mv^2 so we have mgh=2mgh/3 so we have h=2/3H put H = 12 cm h=8cm
A circular saw is powered by a motor. When the saw is used to cut wood, the wood exerts a torque of 0.8 N*m on the saw blade. If the blade rotates with a constant angular velocity of 20 rad/s, the work done on the blade by the motor in 1 min is:
Call the radius of the blade R, and the force at the edge F. Torque = FR = 0.80Nm In 1.0 minute, the total angle of rotation is 20 x 60 = 1200 rad 1 rotation = 2π rad, so the number of rotations = 1200/(2π) This corresponds to a distance = 1200/(2π) x circumference = 1200/(2π) x 2πR = 1200R (If you understand what a radian means (angle subtended by an arc length of 1 radius), you can see that distance =1200R with no calculation). When the saw meets the wood, a force F is applied and the effective distance moved is 1200R. Work done = F x distance = F x 1200R = 1200 FR = 1200 x 0.80 = 960J (Answer C)
The meter stick shown below rotates about an axis through the point marked ., 20 cm from one end. 5 forces act on the stick, one at each end, one at the pivot point, and two 40cm from one end, as shown. The magnitudes of the forces are all the same. Rank the forces according to the magnitudes of the torques they produce about the pivot point, least to greatest
F2 and F5 tie, then F4, then F1, and F3 tie
a wheel starts from rest and has an angular acceleration of 4 rad/s^2. When it has made 10 rev, its angular velocity is:
First, we know that: a = 4 rad/s^2 S = 10 rev = 62.83 rad Now we know that: wf^2-wi^2=2as where wf is the final angular velocity, wi the initial angular velocity, a is the angular acceleration and S the radians. Replacing, we get: Finally, solving for : = 22.41rad/s
A constant torque of 260 acts on a flywheel. If the flywheel makes 25 complete revolutions, how much work has been done by the torque on the flywheel?
For simple linear motion, work = force x distance. W = F x d. For rotation, work = torque x angle. W = τ x θ (angle is in radians). 1 rotation is 2π radians, so 23 rotations = 23 x 2π radians. W = τ x θ = 480 x 23 x 2π = 69,000 J to 2 significant figures.
The rotational inertia of a disk about its axis is 0.7 kg*m^2. When a 2kg weight is added to its rim, 0.4m from the axis, the rotational inertia becomes:
I for a disk= 1/2mr^2 Assuming the new mass is distributed evenly around the periphery at that distance of 0.4 m, then you are just adding 2 kg x (0.4m)^2 to the existing moment of inertia. 0.7+0.32 = 1.02 kg-m^2.
A disk with a rotational inertia of 5kg*m^2 and a radius of 0.25m rotates on a frictionless fixed axis perpendicular to the disk and through its center. A force of 8N is applied along the rotation axis. The angular acceleration of the disk is:
I of disk=1/2mr^2
Two objects are moving in the x, y plane as shown. The magnitude of their total angular momentum (about the origin O) is:
L1=2i*3(3j)=18k L2=j*6(-2i)=12k Ltotal=L1+L2=30kg*m/s^2
A wheel initially has an angular velocity of 36rad/s but after 6s its angular velocity is 24 rad/s. If its angular acceleration is constant, the value is:
Similar to the equation for linear motion, v = vo + at, the equation for angular motion would be: ω = ωo + αt 24 rad/s = 36 rad/s + α(6.0 s) α = -2.0 rad/s^2 (rotation is slowing down)
A 6kg particle moves to the right at 4m/s as shown. The magnitude of its angular momentum about the point O is:
The angle between the tails of the momentum vector and the position vector is 30°; L = m v r sin30 = 6(4)(12)sin30 = 144 kg m2 /s
A certain wheel has a rotational inertia of 12. As it turns through 5 rev its angular velocity increases from 5 rad/s to 6 rad/s. If the net torque is constant its value is:
Torque=I*α I=12 use angular kinematics to find α
A wheel of radius 0.5 rolls w/o sliding on a horizontal surface as shown. Starting from rest, the wheel moves with constant angular acceleration 6 rad/s^2. The distance traveled by the center of the wheel from t=0 to t=3 is:
We know that in case of no slipping, a=rα =(0.5)(6)=3m/s^2 Distance travelled is S=1/2at^2 =13.5m
For a wheel spinning on an axis through its center, the ratio of the radial acceleration of a point on the rim to the radial acceleration of a point halfway between the center of the rim is:
a = α*r where α = angular acceleration a1 = α*r/2 a/a1 = 2
A 16kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter 0.4m and hangs vertically, as shown. The rotational inertia of the flywheel is 0.5. When the block is released and the cord unwinds, the acceleration of the block is:
a=g/1+I/mr^2=5.51 m/s^2
This graph shows the angular velocity of a turntable as a function of time. What is its angular acceleration at t=3.5s?
find derivative at t=3.5s: -10 rad/s^s
A man, with his arms at his sides, is spinning on a light frictionless turntable. When he extends his arms:
his angular momentum remains the same
The torque exerted on an object can be written as T=R*F. Here, r:
is a vector pointing from the axis of rotation to the point where the force is applied
The coefficient of static friction between a certain cylinder and a horizontal floor is 0.4. If the rotational inertia of the cylinder about its symmetry is given by I=1/2mr^2, then the max acceleration the cylinder can have w/o sliding is:
multiply the coefficient of static friction by 2: 0.8g
Two uniform cylinders have different masses and different rotational inertias. They simultaneously start from rest at the top of an inclined plane and roll without sliding down the plane. The cylinder that gets to the bottom first is:
neither (they arrive together)
A child, riding on a large merry-go-round, travels a distance of 3000m in a circle of diameter 40m. The total angle through which she revolves is:
r=d/2=40/2=20m 3000/2πr=23.8rev 23.8*2π=150rad
If a wheel is turning at 3.0 rad/s, the time it takes to complete one revolution is about
t = 2π/3 The time is approximately 2.1 seconds.
A uniform disk, a thin loop, and a uniform solid sphere, all with the same mass and same outer radius, are each free to rotate about a fixed axis through its center. Assume the hoop is connected to the rotation axis by light spokes. With the objects starting from rest, identical forces are simultaneously applied to the rims, as shown. Rank the objects according to their angular velocities after a given time t, least to greatest.
the angular velocity is inversely proportional to the moment of inertia: hoop, disk, sphere
A block with mass M, on the end of a string, moves in a circle on a horizontal frictionless table as shown. As the string is slowly pulled through a small hole in the table:
the kinetic energy of M remains constant
The figure shows a cylinder of radius 0.7m rotating clockwise about its axis at 10 rad/s. The speed of the point P is:
v=wr =7m/s
The fan shown has been turned on and is slowing as it rotates clockwise. The direction of the acceleration of the point X on the fan tip could be:
vector pointing directly to the left: ←
Wrapping paper is being unwrapped from a 5cm radius tube, free to rotate on its axis. If it is pulled at the constant rate of 10cm/s and does not slip on the tube, the angular velocity of the tube is:
w=v/r -convert both values to m w= 2rad/s
A rod is pivoted about its center. A 5N force is applied 4m from the pivot and another 5N force is applied 2m from the pivot, as shown. The magnitude of the total torque about the pivot is:
τ = 4.0×5.0sin 30 + 2.0×5.0sin 30 =15 N
