Mechanics Booster-4

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#Physics #Mechanics #Rotational Dynamics

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9 Answers

1161

Akash Anand ·2014-02-23 22:09:22

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1133

Sourish Ghosh ·2014-02-15 01:13:46

a) 5Î¼g3
b) 3w₀R11Î¼g
c) 5w₀R11Î¼g

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Akash Anand All answers are wrong...try again.
Upvote·0· Reply ·2014-02-15 02:04:08
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486

Niraj kumar Jha ·2014-02-15 03:25:41

a)Î¼g3
b)3w₀R4Î¼g

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Akash Anand Check the second answer again
Upvote·0· Reply ·2014-02-15 03:49:21
Sourish Ghosh I think the 2nd part should be 3/7 instead of 3/4.
Upvote·0· Reply ·2014-02-15 04:42:17
Niraj kumar Jha we should conserve ang. momentum right?
Upvote·0· Reply ·2014-02-15 06:50:32
Niraj kumar Jha is it 9/4for 2nd part
Upvote·0· Reply ·2014-02-17 02:25:54
Akash Anand Sourish is right for the 2nd part.
Upvote·0· Reply ·2014-02-17 03:29:13
Sourish Ghosh Sir, the 3rd part is confusing. It says pure rolling persists after t*. That means after t* the disc will move linearly at a constant speed (are we neglecting friction?) while the block will decelerate (its velocity will change). Now the question says that the block and the disc move together always which implies they will have the same velocity at any instant. How is this possible?
Upvote·0· Reply ·2014-02-17 07:18:23
Soumyadeep Basu Pure rolling does not necessarily mean that linear velocity will remain constant. It just means that v=r.(omega) .
Upvote·0· Reply ·2014-02-17 07:31:11
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51

Aniq Ur Rahman ·2014-02-15 04:49:28

a) Î¼g3

b) 3Ï‰_oRÎ¼g

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Akash Anand Second answer is wrong.
Upvote·0· Reply ·2014-02-15 05:39:14
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51

Aniq Ur Rahman ·2014-02-15 05:34:48

c) √(65)RÏ‰_oÎ¼g

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Akash Anand Wrong ..try again
Upvote·0· Reply ·2014-02-15 05:39:24
Aniq Ur Rahman I guess this is wrong, cuz i applied Conservation of Energy, but energy will be lost due to friction.
Upvote·0· Reply ·2014-02-15 05:42:12
Aniq Ur Rahman I guess this is wrong, cuz i applied Conservation of Energy, but energy will be lost due to friction.
Upvote·0· Reply ·2014-02-15 05:42:16
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1161

Akash Anand ·2014-02-16 23:34:52

Keep trying ..I want answer of this question from you only.

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865

Soumyadeep Basu ·2014-02-17 07:37:36

(a) Î¼g3
(b) 3RÏ‰_o7Î¼g
(c) RÏ‰_o2Î¼g

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865

Soumyadeep Basu ·2014-02-17 07:37:39

(a) Î¼g3
(b) 3RÏ‰_o7Î¼g
(c) RÏ‰_o2Î¼g

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Akash Anand Check the last part again :)
Upvote·0· Reply ·2014-02-18 05:06:42
Shaswata Roy Even I got the same answer for the last one.Can we not equate angular momentum at time t :(omega) - 2(mu)gt/R=0 ?Am I missing out something?
Upvote·0· Reply ·2014-02-23 05:30:28
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1161

Akash Anand ·2014-02-18 22:15:02

The correct answer is:
(a) Î¼g3
(b) 3Ï‰₀R7Î¼g
(c) Ï‰₀RÎ¼g

Can any one post the solution?

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Sourish Ghosh Sir, please post the solution. I am having lots of confusions regarding this problem.
Upvote·0· Reply ·2014-02-23 00:34:24
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Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω