Question

1- A force acting on a machine part is **F** = -4.5
**i** + 3.9 **j** . The vector from the
origin to the point where the force is applied is
**r** = -0.59 **i** + 0.23
**j**. Calculate the magnitude of the vector torque,
in N-m. *Let a quantity along the positive axis be
positive.*

*2- *The flywheel of an engine has moment of
inertia 2.1 kg m^{2} about its rotation axis. What constant
torque is required to bring it up to an angular speed of 406
rev/min in 6.96 s, starting from rest, in N-m?

Answer #1

Force F= (2.84 i - 1.39 k)N acts on a pebble with position
vector r= (6.54 j - 1.60 k)m , relative to the origin. What is the
resulting torque acting on the pebble about (a)
the origin and (b) a point with coordinates (4.96
m, 0, -1.43 m)?

1)A 1kg solid ball (I=2/3mr^2) rolls on a horizontal surface at
the rate of 20m/s. its total kinetic energy is
2)the angular momentum of a particle of mass 0.01kg and position
vector r=(10i+6j)m and moving with a velocity 5i m/s about the
origin is
3)a flywheel is given a speed of 900rpm in 15s. determine the
linear acceleration of a point 0.5m from the horizontal axis is
4)a wheel of radius 0.04 and moment of inertia 3.2x10^-4kgm^2 has a
constant...

An object moving in the xy-plane is subjected to the force F =
2xy i + 3y j N, where x and y are in m.
a. The particle moves from the origin to the point with
coordinates (a, b) by moving first along the x-axis to (a, 0), then
parallel to the y-axis. How much work does the force do?
b. The particle moves from the origin to the point with
coordinates (a, b) by moving first along the...

1) The combination of an applied force and a friction force
produces a constant total torque of 35.8 N · m on a wheel rotating
about a fixed axis. The applied force acts for 5.90 s. During this
time, the angular speed of the wheel increases from 0 to 10.1
rad/s. The applied force is then removed, and the wheel comes to
rest in 60.1 s.
(a) Find the moment of inertia of the wheel.
kg · m2
(b) Find...

A) A man holds a 185-N ball in his hand, with the forearm
horizontal (see the figure). He can support the ball in this
position because of the flexor muscle force , which is applied
perpendicular to the forearm. The forearm weighs 18.2 N and has a
center of gravity as indicated. Find (a) the magnitude of and the
(b) magnitude and (c) direction (as a positive angle
counterclockwise from horizontal) of the force applied by the upper
arm bone...

1) A torque of 1.20 N m is applied to a thin rod of mass 2.50 kg
and length 50.0 cm pivoted about its center and at rest. How fast
is the rod spinning after 4.25 s?
a. 32.6 rad/s
b. 8.16 rad/s
c. 97.9 rad/s
d. 24.5 rad/s
2) A torque of 1.20 N m is applied to a thin rod of mass 2.50 kg
and length 50.0 cm pivoted about one end and at rest. How fast is...

A child pushes her friend (m = 25 kg) located at a radius r =
1.5 m on a merry-go-round (rmgr = 2.0 m, Imgr
= 1000 kg*m2) with a constant force F = 90 N applied
tangentially to the edge of the merry-go-round (i.e., the force is
perpendicular to the radius). The merry-go-round resists spinning
with a frictional force of f = 10 N acting at a radius of
1 m and a frictional torque τ = 15 N*m...

1)A 1.0 kg mass is located at (2.3, 0, -7.9) m and moving at (0,
5.7, 3.9) m/s. What is the angular momentum of this mass about the
origin?
A.
(-3.0, 0.0, -76)
B.
(13, -9.0, 45)
C.
(45, -9.0, 44)
D.
(45, -9.0, 13)
E.
(-76, 0.0, -3.0)
F.
(0.0, 0.0, 0.0)
G.
(-13, 9.0, -45)
H.
(-45, 9.0, -13)
I.
(76, 0.0, 3.0)
J.
(-45, 9.0, -44)
K.
(-44, 9.0, -45)
L.
(44, -9.0, 45)
M.
(3.0, 0.0,...

1. Review your notes from 02-28 and your calculation of
the moment of inertia of the "rigid rotator,"
two masses, 7 kg each
attached to a light titanium bar (neglect this weight)
separation 4.20 m
rotation axis halfway between the two masses, perpendicular to
the bar.
Our calculation of moment of inertia Iwas
I = ∑ left, right ( m r 2 ) = ( 7 k g ) ( 2.10 m ) 2 + ( 7 k g )...

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