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Cinematica dei sistemi di corpi rigidi

Authored by Giovanni Bianchi

Mathematics

University

Used 40+ times

Cinematica dei sistemi di corpi rigidi
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11 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

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Dove si trova, in questo atto di moto, il CIR della biella BC?

A

B

C

\infty

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

Dove si trova in questo atto di moto il CIR della biella BC?

A

B

C

\infty

3.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

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In questo atto di moto...

vC = vB\overrightarrow{v_C\ }\ =\ \overrightarrow{v_B}

vC = 0\overrightarrow{v_C\ }\ =\ 0

vC = vc, maxi\overrightarrow{v_C\ }\ =\ \left|v_{c,\ \max}\right|\overrightarrow{i}

vC = vc, maxi\overrightarrow{v_C\ }\ =\ -\left|v_{c,\ \max}\right|\overrightarrow{i}

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

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In questo atto di moto...

aC = aB\overrightarrow{a_C}\ =\ \overrightarrow{a_B}

aC = 0\overrightarrow{a_C}\ =\ 0

aC = aC,maxi\overrightarrow{a_C}\ =\ \left|a_{C,\max}\right|\overrightarrow{i}

aC = aC,maxi\overrightarrow{a_C}\ =-\ \left|a_{C,\max}\right|\overrightarrow{i}

5.

MULTIPLE SELECT QUESTION

1 min • 1 pt

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Per il sistema rappresentato in figura

La traiettoria di G è una circonferenza

aB = aC\overrightarrow{a_B}\ =\ \overrightarrow{a_C}

ωAB = ωCD\overrightarrow{\omega_{AB}}\ =-\ \overrightarrow{\omega_{CD}}

dωABdt = dωCDdt\frac{\text{d}\overrightarrow{\omega_{AB}}}{\text{d}t}\ =\ \frac{\text{d}\overrightarrow{\omega_{CD}}}{\text{d}t}

6.

MULTIPLE SELECT QUESTION

1 min • 1 pt

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Per il sistema rappresentato in figura, in questo atto di moto...

vB = vC\overrightarrow{v_B\ }\ =\ \overrightarrow{v_C}

dωBCdt = 0\frac{\text{d}\overrightarrow{\omega_{BC}}}{\text{d}t}\ =\ 0

ωAB = ωCD\overrightarrow{\omega_{AB}}\ =\ \overrightarrow{\omega_{CD}}

aB = aC\overrightarrow{a_B}\ =\ \overrightarrow{a_C}

7.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

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Per il sistema rappresentato in figura, considerando un sistema di riferimento relativo traslante con origine in B...

La traiettoria di C è rettilinea

vC,tr (BA)\overrightarrow{v_{C,tr}}\ \ \ \perp\ \left(B-A\right)

vC,tr (CA)\overrightarrow{v_{C,tr}}\ \ \ \perp\ \left(C-A\right)

vC,rel (CB)\overrightarrow{v_{C,rel}\ }\ \parallel\ \left(C-B\right)

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