Physics Problems & Solutions: Torque

Showing posts with label Torque. Show all posts

Electromagnetism - Angular Momentum

Two small pith balls, each carrying a charge q, are attached to the ends of a light rod of length d, which is suspended from the ceiling by a thin torsion-free fiber, as shown in the figure. There is a uniform magnetic field B, pointing straight down, in the cylindrical region of radius R around the fiber. The system is initially at rest. If the magnetic field is turned off, which of the following describes what happens to the system?

It rotates with angular momentum qBR².
It rotates with angular momentum ¼ qBd².
It rotates with angular momentum ½ qBRd.
It does not rotate because to do so would violate conservation of angular momentum.
It does not move because magnetic forces do no work.

(GR9677 #87)

Solution:

B and C are FALSE.
Angular momentum is finite, it cannot go to infinite.
If d → ∞ , angular momentum → ∞

D. FALSE
Since there is external torque, angular momentum is not conserved.

E. FALSE
The system will rotate due to B.

Answer: A:

Calculation:
Faraday's Law: $\varepsilon=-\frac{d \Phi _B}{dt}=\vec{E}\cdot dl$

with Φ = BπR² and dl = 2πd/2 = πd

$\pi R^2 \frac{dB}{dt}=E\pi d\rightarrow E=\frac{\dot{B}R^2}{d}$
$F=qE=\frac{q\dot{B}R^2}{d}$

Torque: $\tau =\sum\vec{r}\times \vec{F}$

Since there is a force contribution from each charge and by the right-hand-rule their cross products with the moment-arm point in the same direction,

$\tau =2 \frac{d}{2}\frac{q\dot{B}R^2}{d}=q\dot{B}R^2$

Torque and Angular momentum:

$\\\sum \tau = \dot{L} \\\rightarrow \dot{L} = q\dot{B}R^2 \\\rightarrow L= qBR^2$

Classical Mechanics - Torque

A solid cone hangs from a frictionless pivot at the origin O as shown above. If î, ĵ and k̂ are unit vectors, and a, b, and c are positive constants, which of the following forces F applied to the rim of the cone at point P results in a torque τ on the cone with a negative component τ_z?

A. F = ak̂, P is (0, b, −c)
B. F = −ak̂, P is (0, −b, −c)
C. F = aĵ, P is (–b, 0, −c)
D. F = aĵ, P is (b, 0, −c)
E. F = −ak̂, P is (−b, 0, −c)

(GR9277 #08)

Solution:

Torque:

Result desired: negative component of τ_zso we are looking for minus k̂ component.

For k̂ component: r_xF_y − r_yF_x

Thus, F_xand F_ycannot be zero

(A) FALSE
F = ak̂ → F_x= 0 and F_y= 0

(B) and (E) FALSE
F = −ak̂ → F_x= 0 and F_y= 0

(C) TRUE
F = aĵ → F_x= 0, F_y= a, r_x= −b, r_y= 0,
P is (–b, 0, −c) → r_x= −b, r_y= 0
r_xF_y − r_yF_x= −ab − 0 = −abk̂ (negative k̂ component)

(D) FALSE
F = aĵ → F_x= 0, F_y= a,
P is (b, 0, −c) → r_x= b, r_y= 0,
→ r_xF_y − r_yF_x= ab − 0 = abk̂ (positive k̂ component)

Answer: C

Classical Mechanics - Rotational Motion

Questions 41-42

A cylinder with moment inertia 4 kgm² about a fixed axis initially rotates at 80 radians per second about this axis. A constant torque is applied to slow it down to 40 radians per second. The kinetic energy lost by the cylinder is

A. 80 J
B. 800 J
C. 4000 J
D. 9600 J
E. 19,200 J

(GR9277 #41)

Solution:

Rotational Kinetic Energy: $T_\text{rot} = \frac{1}{2}I\omega^2$

$\Delta T_\text{rot} = \frac{1}{2}I\left (\omega_f^2-\omega_i^2 \right ) = \frac{1}{2}\cdot 4\left (40^2-80^2 \right ) = -9600 \text{ J}$

Answer: D

Classical Mechanics - Rotational Motion

Questions 41-42

If the cylinder takes 10 seconds to reach 40 radian per second, the magnitude of the applied torque is

A. 80 Nm
B. 40 Nm
C. 32 Nm
D. 16 Nm
E. 8 Nm

(GR9277 #42)

Solution:

$\tau = I\alpha = I \frac{\left ( \omega_f - \omega_i \right )}{t}=4\frac{\left ( 40 - 80 \right )}{10}=-16 \text{ Nm}$

Answer: D

Classical Mechanics - Physical Pendulum

A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, shown above, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of configuration I?

A. (6/5)^½
B. (3/2)^½
C. 6/5
D. 3/2
E. 5/3

(GR9277 #61)

Solution:

Angular frequency for physical pendulum:

ω = (MgL/I)^1/2

Pendulum I
M = 2m
I = mr² + mr² = 2mr²
L = r + r = 2r

ω_I= (2mg2r/2mr²)^1/2= (2g/r)^1/2

Pendulum II
M = 2m
I = mr² + m(r/2)² = (⁵/₄)mr²

L = ½ r + r = (³/₂)r

ω_II=[ 2mg (³/₂)r / (⁵/₄)mr²]^1/2= (12g/5r)^1/2

The ratio:

ω_II/ ω_I= (12g/5r)^1/2/ (2g/r)^1/2= 6/5

Answer: A

Classical Mechanics - Angular Speed

A thin plate of mass M, length L, and width 2d is mounted vertically on a frictionless axle along the z-axis. Initially the object is at rest. It is then tapped with a hammer to provide a torque τ, which produces an angular impulse H about the z-axis of magnitude H = ∫ τ dt. What is the angular speed ω of the plate about the z-axis after the tap?

A. H/2Md²
B. H/Md²
C. 2H/Md²
D. 3H/Md²
E. 4H/Md²

(GR9277 #82)

Solution:

H = ∫ τ dt

Torque: τ = Iα
Angular acceleration: α = ω/t → ω = αt

H = ∫ τ dt = Iα ∫dt = Iαt = Iω

Moment inertia for the plate about the z-axis: I = ¹/₃Md²

H = ¹/₃Md²ω
ω = 3H/Md²

Answer: D

Classical Mechanics - Rotational Motion

A uniform rod of length 10 meters and mass 20 kilograms is balanced on a fulcrum with a 40 kg mass on one end of the rod and 20 kg mass on the other end, as shown above. How far is the fulcrum located from the center of the rod?

A. 0 m
B. 1 m
C. 1.25 m
D. 1.5 m
E. 2 m

(GR9277 #100)

Solution:

∑τ = 0

200(5+d) + 200d - 400(5-d) = 0

1000 + 200d + 200d - 2000 + 200d = 0

800d = 1000

d = 10/8 = 1.25

Answer: C

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