A Lazlo Holics problem

As shown in the diagram, an empty hemisphere with radius R = 0.5 m R = \SI[per-mode=symbol]{0.5}{\meter} is fixed to the top of a cart rolling smoothly on a horizontal ground, such that its cross-section is parallel to the ground.

The cart with mass 2 kg \SI[per-mode=symbol]{2}{\kilo\gram} is initially at rest, and a point-like ball with mass 0.5 kg \SI[per-mode=symbol]{0.5}{\kilo\gram} is dropped into the hemisphere tangentially from a height of h = R h=R above one of its edges. The ball then slides with no friction through the hemisphere.

With what force (in Newtons) will the ball press the hemisphere on its lowest point? Take g = 10 m / s 2 g = \SI[per-mode=symbol]{10}{\meter\per\second\squared} .


The answer is 30.

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2 solutions

Rohit Gupta
Jun 24, 2017

Relevant wiki: Uniform circular motion - Medium

In the absence of friction, no force acts in the horizontal direction on the cart and ball system. Therefore, their momentum must be conserved along the horizontal direction.

Let the speed of the cart be v c v_c and the speed of the ball be v b v_b at the instant the ball reaches the bottommost point of the hemisphere. Thus using the momentum conservation we may write.

m cart v c = m ball v b m_{\text{cart}}v_c = m_{\text{ball}}v_b v b = 4 v c . v_b = 4 v_c.

Again due to no friction on the surface of the hemisphere, we can conserve the mechanical energy, m ball g 2 R = 1 2 m cart v c 2 + 1 2 m ball v b 2 m_{\text{ball}}g2R = \frac{1}{2}m_{\text{cart}}v_c^2 +\frac{1}{2}m_{\text{ball}}v_b^2 Substituting the value of v b , m cart v_b, \, m_{\text{cart}} , and m ball m_{\text{ball}} , we get, v c = 1 , v b = 4 m/s . v_c =1, v_b = 4 \text{m/s}.

Now, in the frame of the cart, the path of the ball is circular and its speed equals v b c = 4 + 1 = 5 m/s v_{bc} = 4+1 =5 \text{m/s} .

Applying the equation of centripetal force at the bottommost point of the hemisphere, N m ball g = m ball v b c 2 R N - m_{\text{ball}}g = m_{\text{ball}} \frac{v_{bc}^2}{R} Substituting the values, we get, N = 30 N . N = \boxed{30 \text{N}}.

It would be worth mentioning here that even though the cart is a non inertial frame \textbf{non inertial frame} , we can still apply the centripetal force equation in the cart frame with ball's velocity relative to cart v b c v_{bc} . This is because even after considering fictitious forces \textbf{fictitious forces} on the ball in cart frame, there is no change in F = m a \sum \textbf{F} = m \textbf{a} equation along vertical axis, since the acceleration of the cart has no component along it, thereby no fictitious forces act in vertical ( -g \textbf{-g} ) direction.

Harsh Poonia - 1 year, 10 months ago
Pedro Bastos
Jun 18, 2017

sorry for the typo

Pedro Bastos - 3 years, 11 months ago

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Why did you take m g 2 R mg 2R , shouldn't it be m g 3 R mg 3R ? Or the problem statement should say h = R h=R instead of h = 2 R h =2R .

Rohit Gupta - 3 years, 11 months ago

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Actually, that h refers to the distance of the cart and the ball, the image may cause alternative visions, now i see it.

Pedro Bastos - 3 years, 11 months ago

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