Practice free Rotational Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
When the moment of force is maximum, then what is the angle between force and position vector of the force ?
$ \tau = Fr sin \theta $ $ ( sin \theta)_ { max} = 1 for \theta = 90 $
A force $ 2 \hat i + 3 \hat j $ acts about an axis at a position vector $ ( \hat j + \hat k )$ from the axis, then what is the torque due to the force about the axis ?
$| \vec \tau | = | \vec r \times \vec F | $
In the experiment of balancing moments, suppose the fulcrum is at the 60 cm mark, and a known mass of 2 kg is used onthe longer arm. The greatest mass of mwhich can be balanced against 2 kg such that the minimum distance of either of the masses from the fulcrum is atleast 10 cm. (Neglect mass of metre scale.) What will be the value of m ?
$ m _{max} = ( 2 kg ) ( {y_{max} \over x_{min}} ) = 12kg $
The wedge is kept below the 60 cm mark on the meter scale. Known masses of 1 kg and 2 kg are hung at the 20 cmand 30 cm mark respectively. Where will a 4 kg mass be hung on the meter scale to balance it ? (Neglect mass of meter scale.)
The anit -clockwise moments due to 1 kg and 2 kg are = (2 kg wt) (60-20) cm + (2kg wt)(60-30) cm = (1 kg wt) (40)cm + (2kg wt ) (30)cm = 100 kg wt cm. The clock wise moment due to $ 4 kg = 4 kg. \omega t \times x cm $ $ \Rightarrow 100 = 4x or x = 25 cm $ So the 4 kg mass must be hung at (60 cm + x) = (60 cm + 25 cm) = 85 cm mark to balance the scale
When a metre scale is balanced above a wedge, 1 kg mass is hung at 10 cm mark and a 2 kg mass is hang at the 85 cmmark. To which mark on the meter scale, the fulcrum be shifted (Neglect mass of meter scale) to balance the scale ?
Balancing moment $ 1x =2 (75 - x) \Rightarrow 3x = 150 or x = 50 cm$ Fulcrum is at 10 cm + 50 cm = 60 cm mark.
Use perpendicular axis theorem
The centre of mass of a systems of two particles is
$ Let R_{cm } is at origin $ $ M \vec R_{cm} = m_1 \vec r_1 + m_2 \vec r_2 $ $ O = m_1 \vec r_1 + m_2 \vec r_2 $ $ - m_1 \vec r_1 = m_2 \vec r_2 $ $ { r_1 \over r _2 } = { m_2 \over m_1 } $ -ve sign ignore as distance
Three particles of the same mass lie in the (X, Y) plane, The (X, Y) coordinates of their positions are (1, 1), (2, 2) and (3, 3) respectively. The (X,Y) coordinates of the centre of mass are
The x and y co.ordinates of centre of mass are $ x = { m_1 x_1 + m_2 x_2 + m_3 x_3 \over m_1 + m_2 + m_3 } $ $ as m_1 = m_2 =m_3 $ $ = { 1 \over 3 } (x_1 + x_2 + x_3 ) = 2 $ similarly for y =2 $ \therefore ( x,y) = (2,2) $
Consider a two-particle system with the particles having masses $ M_1 and M_2 $ . . If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved so as to keep the centre of mass at the same position?
$m_1 x_1 = m_2 x_2 $ $ and m_1 ( x_1 - d ) = m_2 ( x_2 - d ) $ $ \therefore m_1 d = m_2 d ' $ $ \therefore d' = { m_1 d \over m_2 } $
From a uniform circular disc of radius R, a circular disc of radius R/6 and having centre at a distance + R/2 from the centre of the disc is removed. Determine the centre of mass of remaining portion of the disc.
Let mass per unit area of disc = m Mass of disc = M = $ \pi R^2 .m $ Mass of removed disc =M ' = $ \pi \left( { R \over 6 } \right) ^2 .m= { \pi R^2 m \over 36} $ from figure 00' = R/2 $ M \times 0 = M' \times { R \over 2} + ( M - M' ) x $ $ M' x = M' { R \over 2 } + Mx $ $ x = \left( { M' \over M - M' } . { R \over 2 } \right) $
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