A 1.0\text{kg} ball at the <strongtop</strong of a vertical loop of radius 2.0\text{m} moves at 8.0\text{m/s}. Using g \approx 10\text{m/s}^2, what is the magnitude of the normal force from the track on the ball?

22\text{N} At the top, both gravity and normal point toward the center (down): N + m g = m v^2/r \Rightarrow N = (1)(64)/2 - (1)(10) = 32 - 10 = 22\text{N}.

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Acemy

Circular Motion

AP Physics 1· difficulty 3/5

1.0~\text{kg}

A
$42~\text{N}$
B
$32~\text{N}$
C
$22~\text{N}$
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D
$52~\text{N}$

Explanation

N + m g = m v^2/r \Rightarrow N = (1)(64)/2 - (1)(10) = 32 - 10 = 22~\text{N}

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