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A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a speed of 24 m/s, the block hangs vertically down. But when the van maintains this same speed around an unbanked curve (radius = 175m) the block swings toward the outside of the curve, then the string makes an angle theta with the vertical. Find theta.

This question aims to develop a practical understanding of Newton’s laws of motion. It uses the concepts of tension in a string, the weight of a body, and the centripetal/centrifugal force.

Any force acting along a string is called the tension in the string. It is denoted by T. The weight of a body with mass m is given by the following formula:

                                                         w = mg

Where g = 9.8 m/s^2 is the gravitational acceleration. The centripetal force is the force acting toward the center of a circle whenever a body is moving in the circular path. It is mathematically given by the following formula:

\[ F = \dfrac{ m v^2 }{ r } \]

Where $ v $ is the speed of the body while $ r $ is the radius of the circle in which the body is moving.

Expert Answer

During the part of motion where the speed of the van is uniform (constant), the block is hanging vertically downward. In this case, the weight $ w \ = \ m g $ is acting vertically downward. According to Newton’s third law of motion, there is an equal and opposite tension force $ T \ = \ w \ = m g $ must be acting vertically upward to balance the force exerted by the weight. We can say that the system is in equilibrium under such circumstances.

During the part of motion where the van is moving along a circular path of radius $ r \ = \ 175 \ m $ with a speed of $ v \ = \ 24 \ m/s $, this equilibrium is disturbed and the block has moved horizontally towards the outer edge of the curve due to the centrifugal force acting in the horizontal direction.

In this case, the weight $ w \ = \ m g $ acting downward is balanced by the vertical component of tension force $ T cos( \theta ) \ = \ w \ = m g $ and the centrifugal force $ F \ = \ \dfrac{ m v^{ 2 } }{ r } $ is balanced by the horizontal component horizontal component of tension force $ T sin( \theta ) \ = \ F \ = \ \dfrac{ m v^{ 2 } }{ r } $.

So we have two equations:

\[ T cos( \theta ) \ = \ m g \ … \ … \ … \ ( 1 ) \]

\[ T sin( \theta ) \ = \ \dfrac{ m v^{ 2 } }{ r } \ … \ … \ … \ ( 2 ) \]

Dividing equation (1) by equation (2):

\[ \dfrac{ T sin( \theta ) }{ T cos( \theta ) } \ = \ \dfrac{ \dfrac{ m v^{ 2 } }{ r } }{ m g } \]

\[ \Rightarrow \dfrac{ sin( \theta ) }{ cos( \theta ) } \ = \ \dfrac{ v^{ 2 } }{ g r } \]

\[ \Rightarrow tan( \theta ) \ = \ \dfrac{ v^{ 2 } }{ g r } \ … \ … \ … \ ( 3 ) \]

\[ \Rightarrow \theta \ = \ tan^{ -1 } \bigg ( \dfrac{ v^{ 2 } }{ g r } \bigg ) \]

Substituting numerical values:

\[ \theta \ = \ tan^{ -1 } \bigg ( \dfrac{ ( 24 \ m/s )^{ 2 } }{ ( 9.8 \ m/s^2 ) ( 175 \ m ) } \bigg ) \]

\[ \Rightarrow \theta \ = \ tan^{ -1 } ( 0.336 ) \]

\[ \Rightarrow \theta \ = \ 18.55^{ \circ } \]

Numerical Result

\[ \theta \ = \ 18.55^{ \circ } \]

Example

Find the angle theta in the same scenario given above, if the speed was 12 m/s.

Recall equation no. (3):

\[ tan( \theta ) \ = \ \dfrac{ v^{ 2 } }{ g r } \]

\[ \Rightarrow \theta \ = \ tan^{ -1 } \bigg ( \dfrac{ ( 12 \ m/s )^{ 2 } }{ ( 9.8 \ m/s^2 ) ( 175 \ m ) } \bigg ) \]

\[ \Rightarrow \theta \ = \ tan^{ -1 } ( 0.084 ) \]

\[ \Rightarrow \theta \ = \ 4.8^{ \circ } \]

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