Phys101
Term:122
Online HW-Ch10-Lec03
Q1:
A meter stick on a horizontal frictionless table top is pivoted at the 80 cm mark (i.e. the axis
of rotation is at the 80 cm mark). It is initially at rest. A horizontal force F 1 is applied
perpendicularly to the end of the stick at 0 cm, as shown. A second horizontal force F 2 (not
shown) is applied at the 100 cm end of the stick. If the stick does not rotate, which statement
is always TRUE?
A.
B.
C.
D.
E.
|F 2 | > |F 1 | for all orientations of F 2
|F 2 | < |F 1 | for all orientations of F 2
|F 2 | = |F 1 | for all orientations of F 2
|F 2 | > |F 1 | for some orientations of F 2 and |F 2 | < |F 1 | for others
|F 2 | > |F 1 | for some orientations of F 2 and |F 2 | = |F 1 | for others
Ans:
A , Since r2 < r1 ⇒ |F2 | should be larger than |F1 | for all orientations in order
to satify |τ1 | = |τ2 |
Q2:
A disk with a rotational inertia of 5.0 kg /m2 and a radius of 0.25 m rotates on a
frictionless fixed axis perpendicular to the disk and through its center. A force of 8.0
N is applied along the rotation axis. Find the angular acceleration (in rad/s2) of the
disk. (Give your answer in three significant figures form)
Ans:
0 , because the force is applied along the axis of rotation, hence the torque due to this
force is zero, therefore the angular acceleration is also zero (from Newton’s 2nd
law). τ = Iα ; if τ = 0 ⇒ α = 0
KFUPM-Physics Department
1
Phys101
Term:122
Online HW-Ch10-Lec03
Q3:
A 16-kg block is attached to a cord that is wrapped around the rim of a flywheel of
diameter 0.40m and hangs vertically, as shown. The rotational inertia of the flywheel
is 0.50 kg/m2. When the block is released and the cord starts unwinding, find the
acceleration (in m/s2) of the block. (Give your answer in three significant figures
form)
𝐀𝐧𝐬:
Pulley:
Newton’s 2nd law for rotation
τ�⃗net = Iα
�⃗
⇒ TR = Iα, where I = 0.5kg/m2 and α =
⇒ TR = I
Block:
a
a
,⇒ T = I 2
R
R
→ (1)
a
R
�⃗net = ma�⃗
Newton’s 2nd law for translation F
⇒ −T + mg = ma �substitue T from (1)�
−I
a
mg
+ mg = ma ⇒ a =
2
R
m + I/R2
=
KFUPM-Physics Department
16 ∗ 9.8
= 5.50 m/s2
16 + 0.5/0.22
2