Physics Laws of Motion

Get insights from 188 questions on Physics Laws of Motion, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Laws of Motion

Follow Ask Question
188

Questions

0

Discussions

6

Active Users

4

Followers

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

By Conservation of Angular Momentum (COAM) about O, just before and after the collision:
Initial angular momentum L? = mv?
Final angular momentum L? = Iω? = (I_rod + I_block)ω? = (M? ²/3 + m? ²)ω?
L? = L?
mv? = (M? ²/3 + m? ²)ω?
ω? = mv / (M? /3 + m? ) = (16) / (21/3 + 1*1) = 6 / (5/3) = 18/5 rad/s

By Conservation of Total Mechanical Energy (COTME) after collision until it comes to rest:
Initial Energy (at the lowest point) = Rotational K.E. = (1/2)Iω? ²
Final Energy (at the highest point) = Potential Energy = mgh_m + Mg h_M
mgh_m = mg (? -? cosθ)
Mg h_M = Mg (? /2 - (? /2)cosθ)
(1/2) * (M? ²/3 + m? ²) * ω? ² = (mg + Mg/2) *

...more

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

From momentum conservation
mui + 0 = mvj + 3mv'
v' = u/3 I - v/3 j
From kinetic energy conservation 1/2 mu² = 1/2 mv² + 1/2 (3m) (u/3)²+ (v/3)²)
Solving, v = u/√2

New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

mω²acosθ = mgsinθ
ω = √ (gtanθ/a)
y = 4cx²

tanθ = dy/dx = 8xC
(tanθ)? , b = 8aC
ω = √ (g*8aC/a) = 2√ (2gC)

 


New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

FR > mgcosθR
F > mgcosθ
F > mg √ (R²- (R-a)²)/R ⇒ Mg√ (1- (R-a)²/R²)

 

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

mv? *cosθ*2=2m (v? /2)
⇒ cosθ=1/2 ⇒ θ=60°
∴ 2θ=120°

New answer posted

a month ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

µ=tanθ ⇒ 3/4=tanθ ⇒ θ=37°
h = R-Rcosθ = 1-1 (4/5)=0.2m

New answer posted

a month ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

dm (t)/dt = bv²
F_thrust = v ( dm/dt )
Force on satellite = -v (dm (t)/dt)
M (t)a = -v (bv²)
a = - (bv³)/M (t)

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

50.00

For no toppling

F a 2 + b m g a 2

μ a 2 + μ b a 2

0.2 a + 0.4 b 0.5 a

0.4 b 0.3 a

b 3 a 4

b 0.75 a (in limiting case)

But is not possible as maximum value of  can be equal to  only.

100 b a m a x = 50.00

New answer posted

a month ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

T 2 = 100 T 2 = F

F = 100 N

New answer posted

a month ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.